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2x^2-40=11x
We move all terms to the left:
2x^2-40-(11x)=0
a = 2; b = -11; c = -40;
Δ = b2-4ac
Δ = -112-4·2·(-40)
Δ = 441
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{441}=21$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-21}{2*2}=\frac{-10}{4} =-2+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+21}{2*2}=\frac{32}{4} =8 $
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