86-6q-3q^2=q^2/4+10

Solution for 86-6q-3q^2=q^2/4+10 equation:

86-6q-3q^2=q^2/4+10

We move all terms to the left:

86-6q-3q^2-(q^2/4+10)=0

We get rid of parentheses

-3q^2-q^2/4-6q-10+86=0

We multiply all the terms by the denominator


-q^2-3q^2*4-6q*4-10*4+86*4=0

We add all the numbers together, and all the variables

-1q^2-3q^2*4-6q*4+304=0

Wy multiply elements

-1q^2-12q^2-24q+304=0

We add all the numbers together, and all the variables

-13q^2-24q+304=0

a = -13; b = -24; c = +304;
Δ = b2-4ac
Δ = -242-4·(-13)·304
Δ = 16384
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16384}=128$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-128}{2*-13}=\frac{-104}{-26}
=+4
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+128}{2*-13}=\frac{152}{-26}
=-5+11/13
$