50=5t+0.9375t^2

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Solution for 50=5t+0.9375t^2 equation:



50=5t+0.9375t^2
We move all terms to the left:
50-(5t+0.9375t^2)=0
We get rid of parentheses
-0.9375t^2-5t+50=0
a = -0.9375; b = -5; c = +50;
Δ = b2-4ac
Δ = -52-4·(-0.9375)·50
Δ = 212.5
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{-(-5)-\sqrt{212.5}}{2*-0.9375}=\frac{5-\sqrt{212.5}}{-1.875} $
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{212.5}}{2*-0.9375}=\frac{5+\sqrt{212.5}}{-1.875} $

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