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45y^2+16y-16=0
a = 45; b = 16; c = -16;
Δ = b2-4ac
Δ = 162-4·45·(-16)
Δ = 3136
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{3136}=56$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-56}{2*45}=\frac{-72}{90} =-4/5 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+56}{2*45}=\frac{40}{90} =4/9 $
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