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(X)=X^2+3X-5
We move all terms to the left:
(X)-(X^2+3X-5)=0
We get rid of parentheses
-X^2+X-3X+5=0
We add all the numbers together, and all the variables
-1X^2-2X+5=0
a = -1; b = -2; c = +5;
Δ = b2-4ac
Δ = -22-4·(-1)·5
Δ = 24
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}$$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{6}}{2*-1}=\frac{2-2\sqrt{6}}{-2} $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{6}}{2*-1}=\frac{2+2\sqrt{6}}{-2} $
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