(x^2-9x)=(2x^2+14)

Solution for (x^2-9x)=(2x^2+14) equation:

(x^2-9x)=(2x^2+14)

We move all terms to the left:

(x^2-9x)-((2x^2+14))=0

We get rid of parentheses

x^2-9x-((2x^2+14))=0


We calculate terms in parentheses: -((2x^2+14)), so:

(2x^2+14)

We get rid of parentheses

2x^2+14

Back to the equation:

-(2x^2+14)


We get rid of parentheses

x^2-2x^2-9x-14=0

We add all the numbers together, and all the variables

-1x^2-9x-14=0

a = -1; b = -9; c = -14;
Δ = b2-4ac
Δ = -92-4·(-1)·(-14)
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-5}{2*-1}=\frac{4}{-2}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+5}{2*-1}=\frac{14}{-2}
=-7
$