3t+t^2-20=0

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Solution for 3t+t^2-20=0 equation: 



3t+t^2-20=0

a = 1; b = 3; c = -20;
Δ = b2-4ac
Δ = 32-4·1·(-20)
Δ = 89
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{-(3)-\sqrt{89}}{2*1}=\frac{-3-\sqrt{89}}{2}
$


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{89}}{2*1}=\frac{-3+\sqrt{89}}{2}
$

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