(10x-3)=(x^2+12)

Solution for (10x-3)=(x^2+12) equation:

(10x-3)=(x^2+12)

We move all terms to the left:

(10x-3)-((x^2+12))=0

We get rid of parentheses

10x-((x^2+12))-3=0


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

(x^2+12)

We get rid of parentheses

x^2+12

Back to the equation:

-(x^2+12)


We get rid of parentheses

-x^2+10x-12-3=0

We add all the numbers together, and all the variables

-1x^2+10x-15=0

a = -1; b = 10; c = -15;
Δ = b2-4ac
Δ = 102-4·(-1)·(-15)
Δ = 40
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{10}}{2*-1}=\frac{-10-2\sqrt{10}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{10}}{2*-1}=\frac{-10+2\sqrt{10}}{-2}
$