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5t^2-15t-100=0
a = 5; b = -15; c = -100;
Δ = b2-4ac
Δ = -152-4·5·(-100)
Δ = 2225
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2225}=\sqrt{25*89}=\sqrt{25}*\sqrt{89}=5\sqrt{89}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-5\sqrt{89}}{2*5}=\frac{15-5\sqrt{89}}{10} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+5\sqrt{89}}{2*5}=\frac{15+5\sqrt{89}}{10} $
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