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(t-2)(t+5)=t-2(t+5)
We move all terms to the left:
(t-2)(t+5)-(t-2(t+5))=0
We multiply parentheses ..
(+t^2+5t-2t-10)-(t-2(t+5))=0
We calculate terms in parentheses: -(t-2(t+5)), so:We get rid of parentheses
t-2(t+5)
We multiply parentheses
t-2t-10
We add all the numbers together, and all the variables
-1t-10
Back to the equation:
-(-1t-10)
t^2+5t-2t+1t-10+10=0
We add all the numbers together, and all the variables
t^2+4t=0
a = 1; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·1·0
Δ = 16
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{16}=4$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4}{2*1}=\frac{-8}{2} =-4 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*1}=\frac{0}{2} =0 $
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