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3t^2+28t-5=0
a = 3; b = 28; c = -5;
Δ = b2-4ac
Δ = 282-4·3·(-5)
Δ = 844
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{844}=\sqrt{4*211}=\sqrt{4}*\sqrt{211}=2\sqrt{211}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-2\sqrt{211}}{2*3}=\frac{-28-2\sqrt{211}}{6} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+2\sqrt{211}}{2*3}=\frac{-28+2\sqrt{211}}{6} $
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