2x^2+3=7x-4x^2+8

Solution for 2x^2+3=7x-4x^2+8 equation:

2x^2+3=7x-4x^2+8

We move all terms to the left:

2x^2+3-(7x-4x^2+8)=0

We get rid of parentheses

2x^2+4x^2-7x-8+3=0

We add all the numbers together, and all the variables

6x^2-7x-5=0

a = 6; b = -7; c = -5;
Δ = b2-4ac
Δ = -72-4·6·(-5)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-13}{2*6}=\frac{-6}{12}
=-1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+13}{2*6}=\frac{20}{12}
=1+2/3
$