3x^2+7x=5(x+1)

Solution for 3x^2+7x=5(x+1) equation:

3x^2+7x=5(x+1)

We move all terms to the left:

3x^2+7x-(5(x+1))=0


We calculate terms in parentheses: -(5(x+1)), so:

5(x+1)

We multiply parentheses

5x+5

Back to the equation:

-(5x+5)


We get rid of parentheses

3x^2+7x-5x-5=0

We add all the numbers together, and all the variables

3x^2+2x-5=0

a = 3; b = 2; c = -5;
Δ = b2-4ac
Δ = 22-4·3·(-5)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-8}{2*3}=\frac{-10}{6}
=-1+2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+8}{2*3}=\frac{6}{6}
=1
$