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-0.2y^2+45=0
a = -0.2; b = 0; c = +45;
Δ = b2-4ac
Δ = 02-4·(-0.2)·45
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{36}=6$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6}{2*-0.2}=\frac{-6}{-0.4} =+15 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6}{2*-0.2}=\frac{6}{-0.4} =-15 $
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