3x^2+8x-52=0

Solution for 3x^2+8x-52=0 equation:

3x^2+8x-52=0

a = 3; b = 8; c = -52;
Δ = b2-4ac
Δ = 82-4·3·(-52)
Δ = 688
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:

$\sqrt{\Delta}=\sqrt{688}=\sqrt{16*43}=\sqrt{16}*\sqrt{43}=4\sqrt{43}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4\sqrt{43}}{2*3}=\frac{-8-4\sqrt{43}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4\sqrt{43}}{2*3}=\frac{-8+4\sqrt{43}}{6}
$