-x^2+27=(x-4)(x-3)

Solution for -x^2+27=(x-4)(x-3) equation:

-x^2+27=(x-4)(x-3)

We move all terms to the left:

-x^2+27-((x-4)(x-3))=0

We add all the numbers together, and all the variables

-1x^2-((x-4)(x-3))+27=0

We multiply parentheses ..

-1x^2-((+x^2-3x-4x+12))+27=0


We calculate terms in parentheses: -((+x^2-3x-4x+12)), so:

(+x^2-3x-4x+12)

We get rid of parentheses

x^2-3x-4x+12

We add all the numbers together, and all the variables

x^2-7x+12

Back to the equation:

-(x^2-7x+12)


We get rid of parentheses

-1x^2-x^2+7x-12+27=0

We add all the numbers together, and all the variables

-2x^2+7x+15=0

a = -2; b = 7; c = +15;
Δ = b2-4ac
Δ = 72-4·(-2)·15
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*-2}=\frac{-20}{-4}
=+5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+13}{2*-2}=\frac{6}{-4}
=-1+1/2
$