(X^2+6x-5)/(x+3)^2=0

Solution for (X^2+6x-5)/(x+3)^2=0 equation:

(X^2+6X-5)/(X+3)^2=0


Domain of the equation: (X+3)^2!=0

X∈R


We multiply all the terms by the denominator


(X^2+6X-5)=0

We get rid of parentheses

X^2+6X-5=0

a = 1; b = 6; c = -5;
Δ = b2-4ac
Δ = 62-4·1·(-5)
Δ = 56
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:

$\sqrt{\Delta}=\sqrt{56}=\sqrt{4*14}=\sqrt{4}*\sqrt{14}=2\sqrt{14}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{14}}{2*1}=\frac{-6-2\sqrt{14}}{2}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{14}}{2*1}=\frac{-6+2\sqrt{14}}{2}
$