2x^2-19x=15

Solution for 2x^2-19x=15 equation:

2x^2-19x=15

We move all terms to the left:

2x^2-19x-(15)=0

a = 2; b = -19; c = -15;
Δ = b2-4ac
Δ = -192-4·2·(-15)
Δ = 481
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-\sqrt{481}}{2*2}=\frac{19-\sqrt{481}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+\sqrt{481}}{2*2}=\frac{19+\sqrt{481}}{4}
$