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4t^2-8t-7=0
a = 4; b = -8; c = -7;
Δ = b2-4ac
Δ = -82-4·4·(-7)
Δ = 176
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{176}=\sqrt{16*11}=\sqrt{16}*\sqrt{11}=4\sqrt{11}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4\sqrt{11}}{2*4}=\frac{8-4\sqrt{11}}{8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4\sqrt{11}}{2*4}=\frac{8+4\sqrt{11}}{8} $
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