T^2-3t-4=(t+2)(t+7)

Solution for T^2-3t-4=(t+2)(t+7) equation:

^2-3T-4=(T+2)(T+7)

We move all terms to the left:

^2-3T-4-((T+2)(T+7))=0

We add all the numbers together, and all the variables

-3T-((T+2)(T+7))=0

We multiply parentheses ..

-((+T^2+7T+2T+14))-3T=0


We calculate terms in parentheses: -((+T^2+7T+2T+14)), so:

(+T^2+7T+2T+14)

We get rid of parentheses

T^2+7T+2T+14

We add all the numbers together, and all the variables

T^2+9T+14

Back to the equation:

-(T^2+9T+14)


We add all the numbers together, and all the variables

-3T-(T^2+9T+14)=0

We get rid of parentheses

-T^2-3T-9T-14=0

We add all the numbers together, and all the variables

-1T^2-12T-14=0

a = -1; b = -12; c = -14;
Δ = b2-4ac
Δ = -122-4·(-1)·(-14)
Δ = 88
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:

$\sqrt{\Delta}=\sqrt{88}=\sqrt{4*22}=\sqrt{4}*\sqrt{22}=2\sqrt{22}$
$T_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-2\sqrt{22}}{2*-1}=\frac{12-2\sqrt{22}}{-2}
$
$T_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+2\sqrt{22}}{2*-1}=\frac{12+2\sqrt{22}}{-2}
$