5x^2-7x=2(2x^2-4x+3)

Solution for 5x^2-7x=2(2x^2-4x+3) equation:

5x^2-7x=2(2x^2-4x+3)

We move all terms to the left:

5x^2-7x-(2(2x^2-4x+3))=0


We calculate terms in parentheses: -(2(2x^2-4x+3)), so:

2(2x^2-4x+3)

We multiply parentheses

4x^2-8x+6

Back to the equation:

-(4x^2-8x+6)


We get rid of parentheses

5x^2-4x^2-7x+8x-6=0

We add all the numbers together, and all the variables

x^2+x-6=0

a = 1; b = 1; c = -6;
Δ = b2-4ac
Δ = 12-4·1·(-6)
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-5}{2*1}=\frac{-6}{2}
=-3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+5}{2*1}=\frac{4}{2}
=2
$