10x^2+42x=15

Solution for 10x^2+42x=15 equation:

10x^2+42x=15

We move all terms to the left:

10x^2+42x-(15)=0

a = 10; b = 42; c = -15;
Δ = b2-4ac
Δ = 422-4·10·(-15)
Δ = 2364
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:

$\sqrt{\Delta}=\sqrt{2364}=\sqrt{4*591}=\sqrt{4}*\sqrt{591}=2\sqrt{591}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-2\sqrt{591}}{2*10}=\frac{-42-2\sqrt{591}}{20}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+2\sqrt{591}}{2*10}=\frac{-42+2\sqrt{591}}{20}
$