5c^2=31c+28

Solution for 5c^2=31c+28 equation:

5c^2=31c+28

We move all terms to the left:

5c^2-(31c+28)=0

We get rid of parentheses

5c^2-31c-28=0

a = 5; b = -31; c = -28;
Δ = b2-4ac
Δ = -312-4·5·(-28)
Δ = 1521
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1521}=39$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-31)-39}{2*5}=\frac{-8}{10}
=-4/5
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-31)+39}{2*5}=\frac{70}{10}
=7
$