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X^2=36+4X-28
We move all terms to the left:
X^2-(36+4X-28)=0
We add all the numbers together, and all the variables
X^2-(4X+8)=0
We get rid of parentheses
X^2-4X-8=0
a = 1; b = -4; c = -8;
Δ = b2-4ac
Δ = -42-4·1·(-8)
Δ = 48
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}$$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{3}}{2*1}=\frac{4-4\sqrt{3}}{2} $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{3}}{2*1}=\frac{4+4\sqrt{3}}{2} $
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