2x^2-22x=15

Solution for 2x^2-22x=15 equation:

2x^2-22x=15

We move all terms to the left:

2x^2-22x-(15)=0

a = 2; b = -22; c = -15;
Δ = b2-4ac
Δ = -222-4·2·(-15)
Δ = 604
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:

$\sqrt{\Delta}=\sqrt{604}=\sqrt{4*151}=\sqrt{4}*\sqrt{151}=2\sqrt{151}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-2\sqrt{151}}{2*2}=\frac{22-2\sqrt{151}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+2\sqrt{151}}{2*2}=\frac{22+2\sqrt{151}}{4}
$