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16y+2y^2-130=0
a = 2; b = 16; c = -130;
Δ = b2-4ac
Δ = 162-4·2·(-130)
Δ = 1296
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{1296}=36$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-36}{2*2}=\frac{-52}{4} =-13 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+36}{2*2}=\frac{20}{4} =5 $
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