7b^2+31b-16=4

Solution for 7b^2+31b-16=4 equation:

7b^2+31b-16=4

We move all terms to the left:

7b^2+31b-16-(4)=0

We add all the numbers together, and all the variables

7b^2+31b-20=0

a = 7; b = 31; c = -20;
Δ = b2-4ac
Δ = 312-4·7·(-20)
Δ = 1521
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1521}=39$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(31)-39}{2*7}=\frac{-70}{14}
=-5
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(31)+39}{2*7}=\frac{8}{14}
=4/7
$