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2x^2-18x=-40
We move all terms to the left:
2x^2-18x-(-40)=0
We add all the numbers together, and all the variables
2x^2-18x+40=0
a = 2; b = -18; c = +40;
Δ = b2-4ac
Δ = -182-4·2·40
Δ = 4
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{4}=2$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2}{2*2}=\frac{16}{4} =4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2}{2*2}=\frac{20}{4} =5 $
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