The basic quadratic equation is Ax2 + Bx + C = 0

This equations has 2 solutions that satisfy it, x1 and x2.

The mathematical answer is x = ( -B +- SQRT( B2 - 4AC ) ) / 2A

A quasi-geometric answer can be calculated graphically, with a ruler and a compass.

The key to these graphical calculations is a right triangle with 3 sides as follows:

SQRT( ABS( C / A ) )

B / 2A

Solution Circle Radius (SCR) (more on that later)

One of these 3 sides eventually becomes the hypotenuse, and the SCR can be calculated with the Pythagorean Theorem from the other two.

License #1:

If A were 0, the equation would reduce to a linear equation Bx + C = 0, with one solution, x = -C/B

In order to proceed, we accept the limitation that A<>0, and define:

b = B / A

c = C / A

and thus reduce the equation to x2 + bx + c = 0

This is a license because division cannot be accomplished with a ruler and a compass.

We can take that license because the result is the same.

License #2:

After we have divided by A, we are left with x2 + bx + c = 0.

The units of b are the same as the units of x.

The units of c are the same as the units of x2.

In order to operate on a plane, we need the square root of c.

Getting the square root cannot be accomplished with a ruler and a compass.

We take that license so we can operate on a plane with linear magnitudes:

We assume that we are given SQRT(c) as a length.

After having taken these licences, the solution can be truly geometric - and simple!

The Imagination Circle (IC) has as center the origin, and as radius SQRT( ABS( c ) )

Range of c | c<0 | c=0 | c>0 |

Imagination Circle (IC) |

Draw the Solution Circle Center (SCC) on the x-axis, at location -b/2

(If b is positive, -b/2 is negative, so go to the left of the origin)

Range of c | c<0 | c=0 | c>0 |

The Solution Center Circle (SCC)can fall anywhere on the x-axis |

From the Solution Center Circle reach the Imagination Circle

Range of c | c<0 | c=0 | c>0 |

From the Solution Center Circle (SCC)reach the Imagination Circle (IC) |
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SP Notes |
The SP is where the IC meets the y-axis | The SP is the IC | If the SP is inside the IC: Vertically. If outside: Tangentially. |

In all cases we have a right triangle from the 3 points:

The Origin.

The Solution Circle Center (SCC).

The Sizing Point.

The Solution Circle Radius (SCR) starts at the Solution Circle Center (SCC) and goes to the Sizing Point (SP)

The 3 sides of the right triangle are:

The IC-radius (=SQRT(c)).

The distance from the origin to the Solution Circle Center (=b/2).

The Solution Circle Radius (SCR).

Range of c | c<0 | c=0 | c>0 | |

From the Solution Circle Center (SCC)reach the Sizing Point (SP) |
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Calculate the size of the SCR |
SCR
= SQRT( (-b/2)2 + SQRT(-c)2 ) = SQRT( b2/4 - c ) = SQRT( b2/4 - 4c/4 ) = SQRT( b2 - 4c ) / 2 |
SCR = b/2 |
Inside SCR
= SQRT( SQRT(c)2 - (-b/2)2 ) = SQRT( c - b2/4 ) = SQRT( 4c/4 - b2/4 ) = SQRT( 4c – b2 ) / 2 but the units are in i |
Outside SCR
= SQRT( (-b/2)2 - SQRT(-c)2 ) = SQRT( b2/4 - c ) = SQRT( b2/4 - 4c/4 ) = SQRT( b2 - 4c ) / 2 |

Range of c | c<0 | c=0 | c>0 | |

Draw the Solution Circle (SC) |

Range of c | c<0 | c=0 | c>0 | |

Read the solutions on the Solution Circle |
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SolutionsSCR is calculated in Step #4 |
The solutions are real
= -b/2 +- SCR = -b/2 +- SQRT( b2 - 4c ) / 2 = ( -b +- SQRT( b2 - 4c ) ) / 2 x1 = ( -b + SQRT( b2 - 4c ) ) / 2 x2 = ( -b - SQRT( b2 - 4c ) ) / 2 |
The solutions are real
= –b/2 +- SCR = -b/2 +- b/2 x1 = 0 x2 = -b |
The solutions have an imaginary component,
which is symmetric about the x-axis = -b/2 +- SCRi = -b/2 +- ( SQRT( 4c – b2 ) / 2 )i x1 = -b/2 + ( SQRT( 4c – b2 ) / 2 )i x2 = -b/2 - ( SQRT( 4c – b2 ) / 2 )i |
The solutions are real
= -b/2 +- SCR = -b/2 +- SQRT( b2 - 4c ) / 2 = ( -b +- SQRT( b2 - 4c ) ) / 2 x1 = ( -b + SQRT( b2 - 4c ) ) / 2 x2 = ( -b - SQRT( b2 - 4c ) ) / 2 |

With thanks to my brother Greg, for suggesting that such a solution

If you don't have time for all this, just plug in your own numbers: