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4t^2+24t+2=8
We move all terms to the left:
4t^2+24t+2-(8)=0
We add all the numbers together, and all the variables
4t^2+24t-6=0
a = 4; b = 24; c = -6;
Δ = b2-4ac
Δ = 242-4·4·(-6)
Δ = 672
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{672}=\sqrt{16*42}=\sqrt{16}*\sqrt{42}=4\sqrt{42}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-4\sqrt{42}}{2*4}=\frac{-24-4\sqrt{42}}{8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+4\sqrt{42}}{2*4}=\frac{-24+4\sqrt{42}}{8} $
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