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10b^2-15=0
a = 10; b = 0; c = -15;
Δ = b2-4ac
Δ = 02-4·10·(-15)
Δ = 600
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}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{600}=\sqrt{100*6}=\sqrt{100}*\sqrt{6}=10\sqrt{6}$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10\sqrt{6}}{2*10}=\frac{0-10\sqrt{6}}{20} =-\frac{10\sqrt{6}}{20} =-\frac{\sqrt{6}}{2} $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10\sqrt{6}}{2*10}=\frac{0+10\sqrt{6}}{20} =\frac{10\sqrt{6}}{20} =\frac{\sqrt{6}}{2} $
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