7x+2=2(3x-4)x-10

Solution for 7x+2=2(3x-4)x-10 equation:

7x+2=2(3x-4)x-10

We move all terms to the left:

7x+2-(2(3x-4)x-10)=0


We calculate terms in parentheses: -(2(3x-4)x-10), so:

2(3x-4)x-10

We multiply parentheses

6x^2-8x-10

Back to the equation:

-(6x^2-8x-10)


We get rid of parentheses

-6x^2+7x+8x+10+2=0

We add all the numbers together, and all the variables

-6x^2+15x+12=0

a = -6; b = 15; c = +12;
Δ = b2-4ac
Δ = 152-4·(-6)·12
Δ = 513
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{513}=\sqrt{9*57}=\sqrt{9}*\sqrt{57}=3\sqrt{57}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-3\sqrt{57}}{2*-6}=\frac{-15-3\sqrt{57}}{-12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+3\sqrt{57}}{2*-6}=\frac{-15+3\sqrt{57}}{-12}
$