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-3y^2+16y-13=0
a = -3; b = 16; c = -13;
Δ = b2-4ac
Δ = 162-4·(-3)·(-13)
Δ = 100
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{100}=10$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-10}{2*-3}=\frac{-26}{-6} =4+1/3 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+10}{2*-3}=\frac{-6}{-6} =1 $
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