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4y(2)-5+3=30
We move all terms to the left:
4y(2)-5+3-(30)=0
We add all the numbers together, and all the variables
4y^2-32=0
a = 4; b = 0; c = -32;
Δ = b2-4ac
Δ = 02-4·4·(-32)
Δ = 512
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{512}=\sqrt{256*2}=\sqrt{256}*\sqrt{2}=16\sqrt{2}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{2}}{2*4}=\frac{0-16\sqrt{2}}{8} =-\frac{16\sqrt{2}}{8} =-2\sqrt{2} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{2}}{2*4}=\frac{0+16\sqrt{2}}{8} =\frac{16\sqrt{2}}{8} =2\sqrt{2} $
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