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15y*3y=3
We move all terms to the left:
15y*3y-(3)=0
Wy multiply elements
45y^2-3=0
a = 45; b = 0; c = -3;
Δ = b2-4ac
Δ = 02-4·45·(-3)
Δ = 540
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{540}=\sqrt{36*15}=\sqrt{36}*\sqrt{15}=6\sqrt{15}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{15}}{2*45}=\frac{0-6\sqrt{15}}{90} =-\frac{6\sqrt{15}}{90} =-\frac{\sqrt{15}}{15} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{15}}{2*45}=\frac{0+6\sqrt{15}}{90} =\frac{6\sqrt{15}}{90} =\frac{\sqrt{15}}{15} $
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