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20b^2-15=0
a = 20; b = 0; c = -15;
Δ = b2-4ac
Δ = 02-4·20·(-15)
Δ = 1200
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1200}=\sqrt{400*3}=\sqrt{400}*\sqrt{3}=20\sqrt{3}$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20\sqrt{3}}{2*20}=\frac{0-20\sqrt{3}}{40} =-\frac{20\sqrt{3}}{40} =-\frac{\sqrt{3}}{2} $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20\sqrt{3}}{2*20}=\frac{0+20\sqrt{3}}{40} =\frac{20\sqrt{3}}{40} =\frac{\sqrt{3}}{2} $
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