(t+3)(5/9)=40

Solution for (t+3)(5/9)=40 equation:

(t+3)(5/9)=40

We move all terms to the left:

(t+3)(5/9)-(40)=0

We add all the numbers together, and all the variables

(t+3)(+5/9)-40=0

We multiply parentheses ..

(+5t^2+3*5/9)-40=0

We multiply all the terms by the denominator


(+5t^2+3*5-40*9)=0

We get rid of parentheses

5t^2+3*5-40*9=0

We add all the numbers together, and all the variables

5t^2-345=0

a = 5; b = 0; c = -345;
Δ = b2-4ac
Δ = 02-4·5·(-345)
Δ = 6900
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:

$\sqrt{\Delta}=\sqrt{6900}=\sqrt{100*69}=\sqrt{100}*\sqrt{69}=10\sqrt{69}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10\sqrt{69}}{2*5}=\frac{0-10\sqrt{69}}{10}
=-\frac{10\sqrt{69}}{10}
=-\sqrt{69}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10\sqrt{69}}{2*5}=\frac{0+10\sqrt{69}}{10}
=\frac{10\sqrt{69}}{10}
=\sqrt{69}
$