7-4x^2+6x=x^2+7+4x

Solution for 7-4x^2+6x=x^2+7+4x equation:

7-4x^2+6x=x^2+7+4x

We move all terms to the left:

7-4x^2+6x-(x^2+7+4x)=0

We get rid of parentheses

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

We add all the numbers together, and all the variables

-5x^2+2x=0

a = -5; b = 2; c = 0;
Δ = b2-4ac
Δ = 22-4·(-5)·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2}{2*-5}=\frac{-4}{-10}
=2/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2}{2*-5}=\frac{0}{-10}
=0
$