37-(2c+3)=4(c+5)c

Solution for 37-(2c+3)=4(c+5)c equation:

37-(2c+3)=4(c+5)c

We move all terms to the left:

37-(2c+3)-(4(c+5)c)=0

We get rid of parentheses

-2c-(4(c+5)c)-3+37=0


We calculate terms in parentheses: -(4(c+5)c), so:

4(c+5)c

We multiply parentheses

4c^2+20c

Back to the equation:

-(4c^2+20c)


We add all the numbers together, and all the variables

-2c-(4c^2+20c)+34=0

We get rid of parentheses

-4c^2-2c-20c+34=0

We add all the numbers together, and all the variables

-4c^2-22c+34=0

a = -4; b = -22; c = +34;
Δ = b2-4ac
Δ = -222-4·(-4)·34
Δ = 1028
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1028}=\sqrt{4*257}=\sqrt{4}*\sqrt{257}=2\sqrt{257}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-2\sqrt{257}}{2*-4}=\frac{22-2\sqrt{257}}{-8}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+2\sqrt{257}}{2*-4}=\frac{22+2\sqrt{257}}{-8}
$