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4y^2+3y-430=0
a = 4; b = 3; c = -430;
Δ = b2-4ac
Δ = 32-4·4·(-430)
Δ = 6889
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{6889}=83$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-83}{2*4}=\frac{-86}{8} =-10+3/4 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+83}{2*4}=\frac{80}{8} =10 $
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