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2x^2-13x-40=5
We move all terms to the left:
2x^2-13x-40-(5)=0
We add all the numbers together, and all the variables
2x^2-13x-45=0
a = 2; b = -13; c = -45;
Δ = b2-4ac
Δ = -132-4·2·(-45)
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-23}{2*2}=\frac{-10}{4} =-2+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+23}{2*2}=\frac{36}{4} =9 $
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