3x^2+28x-56=0

Solution for 3x^2+28x-56=0 equation:

3x^2+28x-56=0

a = 3; b = 28; c = -56;
Δ = b2-4ac
Δ = 282-4·3·(-56)
Δ = 1456
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:

$\sqrt{\Delta}=\sqrt{1456}=\sqrt{16*91}=\sqrt{16}*\sqrt{91}=4\sqrt{91}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-4\sqrt{91}}{2*3}=\frac{-28-4\sqrt{91}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+4\sqrt{91}}{2*3}=\frac{-28+4\sqrt{91}}{6}
$