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0=2x^2-3x-65
We move all terms to the left:
0-(2x^2-3x-65)=0
We add all the numbers together, and all the variables
-(2x^2-3x-65)=0
We get rid of parentheses
-2x^2+3x+65=0
a = -2; b = 3; c = +65;
Δ = b2-4ac
Δ = 32-4·(-2)·65
Δ = 529
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{529}=23$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-23}{2*-2}=\frac{-26}{-4} =6+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+23}{2*-2}=\frac{20}{-4} =-5 $
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