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5t^2-50=0
a = 5; b = 0; c = -50;
Δ = b2-4ac
Δ = 02-4·5·(-50)
Δ = 1000
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1000}=\sqrt{100*10}=\sqrt{100}*\sqrt{10}=10\sqrt{10}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10\sqrt{10}}{2*5}=\frac{0-10\sqrt{10}}{10} =-\frac{10\sqrt{10}}{10} =-\sqrt{10} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10\sqrt{10}}{2*5}=\frac{0+10\sqrt{10}}{10} =\frac{10\sqrt{10}}{10} =\sqrt{10} $
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