3x^2+13x=100

Solution for 3x^2+13x=100 equation:

3x^2+13x=100

We move all terms to the left:

3x^2+13x-(100)=0

a = 3; b = 13; c = -100;
Δ = b2-4ac
Δ = 132-4·3·(-100)
Δ = 1369
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1369}=37$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-37}{2*3}=\frac{-50}{6}
=-8+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+37}{2*3}=\frac{24}{6}
=4
$