-2t^2+7t+4=0

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Solution for -2t^2+7t+4=0 equation: 



-2t^2+7t+4=0

a = -2; b = 7; c = +4;
Δ = b2-4ac
Δ = 72-4·(-2)·4
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$


$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-9}{2*-2}=\frac{-16}{-4}
=+4
$


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+9}{2*-2}=\frac{2}{-4}
=-1/2
$

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