0=47-4t-5t^2

Solution for 0=47-4t-5t^2 equation:

0=47-4t-5t^2

We move all terms to the left:

0-(47-4t-5t^2)=0

We add all the numbers together, and all the variables

-(47-4t-5t^2)=0

We get rid of parentheses

5t^2+4t-47=0

a = 5; b = 4; c = -47;
Δ = b2-4ac
Δ = 42-4·5·(-47)
Δ = 956
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{956}=\sqrt{4*239}=\sqrt{4}*\sqrt{239}=2\sqrt{239}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{239}}{2*5}=\frac{-4-2\sqrt{239}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{239}}{2*5}=\frac{-4+2\sqrt{239}}{10}
$