-7b^2+53b+24=0

Solution for -7b^2+53b+24=0 equation:

-7b^2+53b+24=0

a = -7; b = 53; c = +24;
Δ = b2-4ac
Δ = 532-4·(-7)·24
Δ = 3481
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3481}=59$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(53)-59}{2*-7}=\frac{-112}{-14}
=+8
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(53)+59}{2*-7}=\frac{6}{-14}
=-3/7
$