Question

$\ \textless \ br /\ \textgreater \ 4t {}^{2} - 64 = 0\ \textless \ br /\ \textgreater \ ​$
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Answer 1

Step-by-step explanation:

4t^2=64

We move all terms to the left:

4t^2-(64)=0

a = 4; b = 0; c = -64;

Δ = b2-4ac

Δ = 02-4·4·(-64)

Δ = 1024

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{1024}=32

t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-32}{2*4}=\frac{-32}{8}

=-4

t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+32}{2*4}=\frac{32}{8}

=4

Answer 2

Step-by-step explanation:

this question's answer is t = 4 , -4