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(6)=x^2+3x-18
We move all terms to the left:
(6)-(x^2+3x-18)=0
We get rid of parentheses
-x^2-3x+18+6=0
We add all the numbers together, and all the variables
-1x^2-3x+24=0
a = -1; b = -3; c = +24;
Δ = b2-4ac
Δ = -32-4·(-1)·24
Δ = 105
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{105}}{2*-1}=\frac{3-\sqrt{105}}{-2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{105}}{2*-1}=\frac{3+\sqrt{105}}{-2} $
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