5t^2-47t+18=0

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Solution for 5t^2-47t+18=0 equation: 



5t^2-47t+18=0

a = 5; b = -47; c = +18;
Δ = b2-4ac
Δ = -472-4·5·18
Δ = 1849
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1849}=43$


$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-47)-43}{2*5}=\frac{4}{10}
=2/5
$


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-47)+43}{2*5}=\frac{90}{10}
=9
$

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