(3T-8)+((9+T))x2=58

Solution for (3T-8)+((9+T))x2=58 equation:

(3-8)+((9+))T2=58

We move all terms to the left:

(3-8)+((9+))T2-(58)=0

determiningTheFunctionDomain
((9+))T2-58+(3-8)=0

We add all the numbers together, and all the variables

(0)T2-58+(-5)=0

We add all the numbers together, and all the variables

T^2-63=0

a = 1; b = 0; c = -63;
Δ = b2-4ac
Δ = 02-4·1·(-63)
Δ = 252
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:

$\sqrt{\Delta}=\sqrt{252}=\sqrt{36*7}=\sqrt{36}*\sqrt{7}=6\sqrt{7}$
$T_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{7}}{2*1}=\frac{0-6\sqrt{7}}{2}
=-\frac{6\sqrt{7}}{2}
=-3\sqrt{7}
$
$T_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{7}}{2*1}=\frac{0+6\sqrt{7}}{2}
=\frac{6\sqrt{7}}{2}
=3\sqrt{7}
$