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0=x^2-4x-96
We move all terms to the left:
0-(x^2-4x-96)=0
We add all the numbers together, and all the variables
-(x^2-4x-96)=0
We get rid of parentheses
-x^2+4x+96=0
We add all the numbers together, and all the variables
-1x^2+4x+96=0
a = -1; b = 4; c = +96;
Δ = b2-4ac
Δ = 42-4·(-1)·96
Δ = 400
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{400}=20$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-20}{2*-1}=\frac{-24}{-2} =+12 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+20}{2*-1}=\frac{16}{-2} =-8 $
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