4t+2t+t+t^2=

Solution for 4t+2t+t+t^2= equation:

4t+2t+t+t^2=

We move all terms to the left:

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

We add all the numbers together, and all the variables

t^2+7t=0

a = 1; b = 7; c = 0;
Δ = b2-4ac
Δ = 72-4·1·0
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-7}{2*1}=\frac{-14}{2}
=-7
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+7}{2*1}=\frac{0}{2}
=0
$