(X+2)(x+5)=x^+7x+10

Solution for (X+2)(x+5)=x^+7x+10 equation:

(X+2)(X+5)=X^+7X+10

We move all terms to the left:

(X+2)(X+5)-(X^+7X+10)=0

We get rid of parentheses

(X+2)(X+5)-X^-7X-10=0

We multiply parentheses ..

(+X^2+5X+2X+10)-X^-7X-10=0

We add all the numbers together, and all the variables

(+X^2+5X+2X+10)-8X-10=0

We get rid of parentheses

X^2+5X+2X-8X+10-10=0

We add all the numbers together, and all the variables

X^2-1X=0

a = 1; b = -1; c = 0;
Δ = b2-4ac
Δ = -12-4·1·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-1}{2*1}=\frac{0}{2}
=0
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+1}{2*1}=\frac{2}{2}
=1
$