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X^2+4X+5=-6X+8
We move all terms to the left:
X^2+4X+5-(-6X+8)=0
We get rid of parentheses
X^2+4X+6X-8+5=0
We add all the numbers together, and all the variables
X^2+10X-3=0
a = 1; b = 10; c = -3;
Δ = b2-4ac
Δ = 102-4·1·(-3)
Δ = 112
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}$$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{112}=\sqrt{16*7}=\sqrt{16}*\sqrt{7}=4\sqrt{7}$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-4\sqrt{7}}{2*1}=\frac{-10-4\sqrt{7}}{2} $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+4\sqrt{7}}{2*1}=\frac{-10+4\sqrt{7}}{2} $
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