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-5t^2+10t+3=0
a = -5; b = 10; c = +3;
Δ = b2-4ac
Δ = 102-4·(-5)·3
Δ = 160
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-4\sqrt{10}}{2*-5}=\frac{-10-4\sqrt{10}}{-10} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+4\sqrt{10}}{2*-5}=\frac{-10+4\sqrt{10}}{-10} $
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