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2y^2-48y+270=0
a = 2; b = -48; c = +270;
Δ = b2-4ac
Δ = -482-4·2·270
Δ = 144
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{144}=12$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-12}{2*2}=\frac{36}{4} =9 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+12}{2*2}=\frac{60}{4} =15 $
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