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-6t^2+8t+7=0
a = -6; b = 8; c = +7;
Δ = b2-4ac
Δ = 82-4·(-6)·7
Δ = 232
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{232}=\sqrt{4*58}=\sqrt{4}*\sqrt{58}=2\sqrt{58}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{58}}{2*-6}=\frac{-8-2\sqrt{58}}{-12} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{58}}{2*-6}=\frac{-8+2\sqrt{58}}{-12} $
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