(6x^2+3x-2)=2(2x^2-x+5)

Solution for (6x^2+3x-2)=2(2x^2-x+5) equation:

(6x^2+3x-2)=2(2x^2-x+5)

We move all terms to the left:

(6x^2+3x-2)-(2(2x^2-x+5))=0

We get rid of parentheses

6x^2+3x-(2(2x^2-x+5))-2=0


We calculate terms in parentheses: -(2(2x^2-x+5)), so:

2(2x^2-x+5)

We multiply parentheses

4x^2-2x+10

Back to the equation:

-(4x^2-2x+10)


We get rid of parentheses

6x^2-4x^2+3x+2x-10-2=0

We add all the numbers together, and all the variables

2x^2+5x-12=0

a = 2; b = 5; c = -12;
Δ = b2-4ac
Δ = 52-4·2·(-12)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-11}{2*2}=\frac{-16}{4}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+11}{2*2}=\frac{6}{4}
=1+1/2
$