4x^2+10x-57=0

Solution for 4x^2+10x-57=0 equation:

4x^2+10x-57=0

a = 4; b = 10; c = -57;
Δ = b2-4ac
Δ = 102-4·4·(-57)
Δ = 1012
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:

$\sqrt{\Delta}=\sqrt{1012}=\sqrt{4*253}=\sqrt{4}*\sqrt{253}=2\sqrt{253}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{253}}{2*4}=\frac{-10-2\sqrt{253}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{253}}{2*4}=\frac{-10+2\sqrt{253}}{8}
$