13x^2+20x+8=100

Solution for 13x^2+20x+8=100 equation:

13x^2+20x+8=100

We move all terms to the left:

13x^2+20x+8-(100)=0

We add all the numbers together, and all the variables

13x^2+20x-92=0

a = 13; b = 20; c = -92;
Δ = b2-4ac
Δ = 202-4·13·(-92)
Δ = 5184
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{5184}=72$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-72}{2*13}=\frac{-92}{26}
=-3+7/13
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+72}{2*13}=\frac{52}{26}
=2
$