10x^2+4x=12

Solution for 10x^2+4x=12 equation:

10x^2+4x=12

We move all terms to the left:

10x^2+4x-(12)=0

a = 10; b = 4; c = -12;
Δ = b2-4ac
Δ = 42-4·10·(-12)
Δ = 496
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:

$\sqrt{\Delta}=\sqrt{496}=\sqrt{16*31}=\sqrt{16}*\sqrt{31}=4\sqrt{31}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{31}}{2*10}=\frac{-4-4\sqrt{31}}{20}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{31}}{2*10}=\frac{-4+4\sqrt{31}}{20}
$