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0=2x^2-16x-18
We move all terms to the left:
0-(2x^2-16x-18)=0
We add all the numbers together, and all the variables
-(2x^2-16x-18)=0
We get rid of parentheses
-2x^2+16x+18=0
a = -2; b = 16; c = +18;
Δ = b2-4ac
Δ = 162-4·(-2)·18
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-20}{2*-2}=\frac{-36}{-4} =+9 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+20}{2*-2}=\frac{4}{-4} =-1 $
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