20x^2+20x-71=9

Solution for 20x^2+20x-71=9 equation:

20x^2+20x-71=9

We move all terms to the left:

20x^2+20x-71-(9)=0

We add all the numbers together, and all the variables

20x^2+20x-80=0

a = 20; b = 20; c = -80;
Δ = b2-4ac
Δ = 202-4·20·(-80)
Δ = 6800
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{6800}=\sqrt{400*17}=\sqrt{400}*\sqrt{17}=20\sqrt{17}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-20\sqrt{17}}{2*20}=\frac{-20-20\sqrt{17}}{40}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+20\sqrt{17}}{2*20}=\frac{-20+20\sqrt{17}}{40}
$