5(2c+7)=c^2+17c+7

Solution for 5(2c+7)=c^2+17c+7 equation:

5(2c+7)=c^2+17c+7

We move all terms to the left:

5(2c+7)-(c^2+17c+7)=0

We multiply parentheses

10c-(c^2+17c+7)+35=0

We get rid of parentheses

-c^2+10c-17c-7+35=0

We add all the numbers together, and all the variables

-1c^2-7c+28=0

a = -1; b = -7; c = +28;
Δ = b2-4ac
Δ = -72-4·(-1)·28
Δ = 161
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-\sqrt{161}}{2*-1}=\frac{7-\sqrt{161}}{-2}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+\sqrt{161}}{2*-1}=\frac{7+\sqrt{161}}{-2}
$