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=12+-3x+-1x^2
We move all terms to the left:
-(12+-3x+-1x^2)=0
We use the square of the difference formula
-(12-3x-1x^2)=0
We get rid of parentheses
1x^2+3x-12=0
We add all the numbers together, and all the variables
x^2+3x-12=0
a = 1; b = 3; c = -12;
Δ = b2-4ac
Δ = 32-4·1·(-12)
Δ = 57
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{57}}{2*1}=\frac{-3-\sqrt{57}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{57}}{2*1}=\frac{-3+\sqrt{57}}{2} $
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