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4y^2-3y=9
We move all terms to the left:
4y^2-3y-(9)=0
a = 4; b = -3; c = -9;
Δ = b2-4ac
Δ = -32-4·4·(-9)
Δ = 153
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{153}=\sqrt{9*17}=\sqrt{9}*\sqrt{17}=3\sqrt{17}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3\sqrt{17}}{2*4}=\frac{3-3\sqrt{17}}{8} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3\sqrt{17}}{2*4}=\frac{3+3\sqrt{17}}{8} $
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