(x^2-7x+12)=(4x-5)

Solution for (x^2-7x+12)=(4x-5) equation:

(x^2-7x+12)=(4x-5)

We move all terms to the left:

(x^2-7x+12)-((4x-5))=0

We get rid of parentheses

x^2-7x-((4x-5))+12=0


We calculate terms in parentheses: -((4x-5)), so:

(4x-5)

We get rid of parentheses

4x-5

Back to the equation:

-(4x-5)


We get rid of parentheses

x^2-7x-4x+5+12=0

We add all the numbers together, and all the variables

x^2-11x+17=0

a = 1; b = -11; c = +17;
Δ = b2-4ac
Δ = -112-4·1·17
Δ = 53
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-\sqrt{53}}{2*1}=\frac{11-\sqrt{53}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+\sqrt{53}}{2*1}=\frac{11+\sqrt{53}}{2}
$