5b^2+20b-57=0

Solution for 5b^2+20b-57=0 equation:

5b^2+20b-57=0

a = 5; b = 20; c = -57;
Δ = b2-4ac
Δ = 202-4·5·(-57)
Δ = 1540
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1540}=\sqrt{4*385}=\sqrt{4}*\sqrt{385}=2\sqrt{385}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-2\sqrt{385}}{2*5}=\frac{-20-2\sqrt{385}}{10}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+2\sqrt{385}}{2*5}=\frac{-20+2\sqrt{385}}{10}
$