6x^2-96=x^2+9x+20

Solution for 6x^2-96=x^2+9x+20 equation:

6x^2-96=x^2+9x+20

We move all terms to the left:

6x^2-96-(x^2+9x+20)=0

We get rid of parentheses

6x^2-x^2-9x-20-96=0

We add all the numbers together, and all the variables

5x^2-9x-116=0

a = 5; b = -9; c = -116;
Δ = b2-4ac
Δ = -92-4·5·(-116)
Δ = 2401
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2401}=49$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-49}{2*5}=\frac{-40}{10}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+49}{2*5}=\frac{58}{10}
=5+4/5
$