I(t)=-16t^2+32t+10

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Solution for I(t)=-16t^2+32t+10 equation:



(I)=-16I^2+32I+10
We move all terms to the left:
(I)-(-16I^2+32I+10)=0
We get rid of parentheses
16I^2-32I+I-10=0
We add all the numbers together, and all the variables
16I^2-31I-10=0
a = 16; b = -31; c = -10;
Δ = b2-4ac
Δ = -312-4·16·(-10)
Δ = 1601
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$I_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$I_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$I_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-31)-\sqrt{1601}}{2*16}=\frac{31-\sqrt{1601}}{32} $
$I_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-31)+\sqrt{1601}}{2*16}=\frac{31+\sqrt{1601}}{32} $

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