(x+10)(x-5)=(x+)

Solution for (x+10)(x-5)=(x+) equation:

(x+10)(x-5)=(x+)

We move all terms to the left:

(x+10)(x-5)-((x+))=0

We add all the numbers together, and all the variables

(x+10)(x-5)-((+x))=0

We multiply parentheses ..

(+x^2-5x+10x-50)-((+x))=0


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

(+x)

We get rid of parentheses

x

Back to the equation:

-(x)


We add all the numbers together, and all the variables

(+x^2-5x+10x-50)-1x=0

We get rid of parentheses

x^2-5x+10x-1x-50=0

We add all the numbers together, and all the variables

x^2+4x-50=0

a = 1; b = 4; c = -50;
Δ = b2-4ac
Δ = 42-4·1·(-50)
Δ = 216
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:

$\sqrt{\Delta}=\sqrt{216}=\sqrt{36*6}=\sqrt{36}*\sqrt{6}=6\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-6\sqrt{6}}{2*1}=\frac{-4-6\sqrt{6}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+6\sqrt{6}}{2*1}=\frac{-4+6\sqrt{6}}{2}
$