-x^2+4x-7=2x^2+9x-5

Solution for -x^2+4x-7=2x^2+9x-5 equation:

-x^2+4x-7=2x^2+9x-5

We move all terms to the left:

-x^2+4x-7-(2x^2+9x-5)=0

We add all the numbers together, and all the variables

-1x^2+4x-(2x^2+9x-5)-7=0

We get rid of parentheses

-1x^2-2x^2+4x-9x+5-7=0

We add all the numbers together, and all the variables

-3x^2-5x-2=0

a = -3; b = -5; c = -2;
Δ = b2-4ac
Δ = -52-4·(-3)·(-2)
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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