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40=10x^2
We move all terms to the left:
40-(10x^2)=0
a = -10; b = 0; c = +40;
Δ = b2-4ac
Δ = 02-4·(-10)·40
Δ = 1600
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{1600}=40$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-40}{2*-10}=\frac{-40}{-20} =+2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+40}{2*-10}=\frac{40}{-20} =-2 $
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