7t(t-5)=3(5-t)

Solution for 7t(t-5)=3(5-t) equation:

7t(t-5)=3(5-t)

We move all terms to the left:

7t(t-5)-(3(5-t))=0

We add all the numbers together, and all the variables

7t(t-5)-(3(-1t+5))=0

We multiply parentheses

7t^2-35t-(3(-1t+5))=0


We calculate terms in parentheses: -(3(-1t+5)), so:

3(-1t+5)

We multiply parentheses

-3t+15

Back to the equation:

-(-3t+15)


We get rid of parentheses

7t^2-35t+3t-15=0

We add all the numbers together, and all the variables

7t^2-32t-15=0

a = 7; b = -32; c = -15;
Δ = b2-4ac
Δ = -322-4·7·(-15)
Δ = 1444
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1444}=38$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-38}{2*7}=\frac{-6}{14}
=-3/7
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+38}{2*7}=\frac{70}{14}
=5
$