-2t^2+8t+7=0

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Solution for -2t^2+8t+7=0 equation:



-2t^2+8t+7=0
a = -2; b = 8; c = +7;
Δ = b2-4ac
Δ = 82-4·(-2)·7
Δ = 120
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:
$\sqrt{\Delta}=\sqrt{120}=\sqrt{4*30}=\sqrt{4}*\sqrt{30}=2\sqrt{30}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{30}}{2*-2}=\frac{-8-2\sqrt{30}}{-4} $
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{30}}{2*-2}=\frac{-8+2\sqrt{30}}{-4} $

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