5t-36+t^2=0

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Solution for 5t-36+t^2=0 equation: 



5t-36+t^2=0

a = 1; b = 5; c = -36;
Δ = b2-4ac
Δ = 52-4·1·(-36)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$


$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-13}{2*1}=\frac{-18}{2}
=-9
$


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+13}{2*1}=\frac{8}{2}
=4
$

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