3x^2+10x-53=0

Solution for 3x^2+10x-53=0 equation:

3x^2+10x-53=0

a = 3; b = 10; c = -53;
Δ = b2-4ac
Δ = 102-4·3·(-53)
Δ = 736
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:

$\sqrt{\Delta}=\sqrt{736}=\sqrt{16*46}=\sqrt{16}*\sqrt{46}=4\sqrt{46}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-4\sqrt{46}}{2*3}=\frac{-10-4\sqrt{46}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+4\sqrt{46}}{2*3}=\frac{-10+4\sqrt{46}}{6}
$