-x^2+4=-7x^2+33x-41

Solution for -x^2+4=-7x^2+33x-41 equation:

-x^2+4=-7x^2+33x-41

We move all terms to the left:

-x^2+4-(-7x^2+33x-41)=0

We add all the numbers together, and all the variables

-1x^2-(-7x^2+33x-41)+4=0

We get rid of parentheses

-1x^2+7x^2-33x+41+4=0

We add all the numbers together, and all the variables

6x^2-33x+45=0

a = 6; b = -33; c = +45;
Δ = b2-4ac
Δ = -332-4·6·45
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-33)-3}{2*6}=\frac{30}{12}
=2+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-33)+3}{2*6}=\frac{36}{12}
=3
$