3t^2+14t+16=-

Solution for 3t^2+14t+16=- equation:

3t^2+14t+16=-

We move all terms to the left:

3t^2+14t+16-(-)=0

We add all the numbers together, and all the variables

3t^2+14t+16-0=0

We add all the numbers together, and all the variables

3t^2+14t+16=0

a = 3; b = 14; c = +16;
Δ = b2-4ac
Δ = 142-4·3·16
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2}{2*3}=\frac{-16}{6}
=-2+2/3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2}{2*3}=\frac{-12}{6}
=-2
$