5(2c+7)-3c=7c(c+5)

Solution for 5(2c+7)-3c=7c(c+5) equation:

5(2c+7)-3c=7c(c+5)

We move all terms to the left:

5(2c+7)-3c-(7c(c+5))=0

We add all the numbers together, and all the variables

-3c+5(2c+7)-(7c(c+5))=0

We multiply parentheses

-3c+10c-(7c(c+5))+35=0


We calculate terms in parentheses: -(7c(c+5)), so:

7c(c+5)

We multiply parentheses

7c^2+35c

Back to the equation:

-(7c^2+35c)


We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-7c^2-28c+35=0

a = -7; b = -28; c = +35;
Δ = b2-4ac
Δ = -282-4·(-7)·35
Δ = 1764
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1764}=42$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-42}{2*-7}=\frac{-14}{-14}
=1
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+42}{2*-7}=\frac{70}{-14}
=-5
$