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5t^2=7
We move all terms to the left:
5t^2-(7)=0
a = 5; b = 0; c = -7;
Δ = b2-4ac
Δ = 02-4·5·(-7)
Δ = 140
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{140}=\sqrt{4*35}=\sqrt{4}*\sqrt{35}=2\sqrt{35}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{35}}{2*5}=\frac{0-2\sqrt{35}}{10} =-\frac{2\sqrt{35}}{10} =-\frac{\sqrt{35}}{5} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{35}}{2*5}=\frac{0+2\sqrt{35}}{10} =\frac{2\sqrt{35}}{10} =\frac{\sqrt{35}}{5} $
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