2(t+1)(t+2)-4(t-1)2=0

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Solution for 2(t+1)(t+2)-4(t-1)2=0 equation: 



2(t+1)(t+2)-4(t-1)2=0

We multiply parentheses

2(t+1)(t+2)-8t+8=0

We multiply parentheses ..

2(+t^2+2t+t+2)-8t+8=0

We multiply parentheses

2t^2+4t+2t-8t+4+8=0

We add all the numbers together, and all the variables

2t^2-2t+12=0

a = 2; b = -2; c = +12;
Δ = b2-4ac
Δ = -22-4·2·12
Δ = -92
Delta is less than zero, so there is no solution for the equation

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