(x2+3x-6)=(x+4)

Solution for (x2+3x-6)=(x+4) equation:

(x2+3x-6)=(x+4)

We move all terms to the left:

(x2+3x-6)-((x+4))=0

We add all the numbers together, and all the variables

(+x^2+3x-6)-((x+4))=0

We get rid of parentheses

x^2+3x-((x+4))-6=0


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

(x+4)

We get rid of parentheses

x+4

Back to the equation:

-(x+4)


We get rid of parentheses

x^2+3x-x-4-6=0

We add all the numbers together, and all the variables

x^2+2x-10=0

a = 1; b = 2; c = -10;
Δ = b2-4ac
Δ = 22-4·1·(-10)
Δ = 44
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:

$\sqrt{\Delta}=\sqrt{44}=\sqrt{4*11}=\sqrt{4}*\sqrt{11}=2\sqrt{11}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{11}}{2*1}=\frac{-2-2\sqrt{11}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{11}}{2*1}=\frac{-2+2\sqrt{11}}{2}
$