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5t^2-5t-210=0
a = 5; b = -5; c = -210;
Δ = b2-4ac
Δ = -52-4·5·(-210)
Δ = 4225
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{4225}=65$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-65}{2*5}=\frac{-60}{10} =-6 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+65}{2*5}=\frac{70}{10} =7 $
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