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 = 17
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-\sqrt{17}}{2*2}=\frac{-7-\sqrt{17}}{4}
$


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{17}}{2*2}=\frac{-7+\sqrt{17}}{4}
$

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