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0=x^2-10x-96
We move all terms to the left:
0-(x^2-10x-96)=0
We add all the numbers together, and all the variables
-(x^2-10x-96)=0
We get rid of parentheses
-x^2+10x+96=0
We add all the numbers together, and all the variables
-1x^2+10x+96=0
a = -1; b = 10; c = +96;
Δ = b2-4ac
Δ = 102-4·(-1)·96
Δ = 484
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{484}=22$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-22}{2*-1}=\frac{-32}{-2} =+16 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+22}{2*-1}=\frac{12}{-2} =-6 $
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