4t2+7t+3=0

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Solution for 4t2+7t+3=0 equation: 



4t^2+7t+3=0

a = 4; b = 7; c = +3;
Δ = b2-4ac
Δ = 72-4·4·3
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$


$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-1}{2*4}=\frac{-8}{8}
=-1
$


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+1}{2*4}=\frac{-6}{8}
=-3/4
$

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