(5c+7)(2c-3)=-2c(c+15)-35

Solution for (5c+7)(2c-3)=-2c(c+15)-35 equation:

(5c+7)(2c-3)=-2c(c+15)-35

We move all terms to the left:

(5c+7)(2c-3)-(-2c(c+15)-35)=0

We multiply parentheses ..

(+10c^2-15c+14c-21)-(-2c(c+15)-35)=0


We calculate terms in parentheses: -(-2c(c+15)-35), so:

-2c(c+15)-35

We multiply parentheses

-2c^2-30c-35

Back to the equation:

-(-2c^2-30c-35)


We get rid of parentheses

10c^2+2c^2-15c+14c+30c-21+35=0

We add all the numbers together, and all the variables

12c^2+29c+14=0

a = 12; b = 29; c = +14;
Δ = b2-4ac
Δ = 292-4·12·14
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-13}{2*12}=\frac{-42}{24}
=-1+3/4
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+13}{2*12}=\frac{-16}{24}
=-2/3
$