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3t^2+6t-35=0
a = 3; b = 6; c = -35;
Δ = b2-4ac
Δ = 62-4·3·(-35)
Δ = 456
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{456}=\sqrt{4*114}=\sqrt{4}*\sqrt{114}=2\sqrt{114}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{114}}{2*3}=\frac{-6-2\sqrt{114}}{6} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{114}}{2*3}=\frac{-6+2\sqrt{114}}{6} $
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