1(t-6)7t=4(2t+3)-10

Solution for 1(t-6)7t=4(2t+3)-10 equation:

1(t-6)7t=4(2t+3)-10

We move all terms to the left:

1(t-6)7t-(4(2t+3)-10)=0

We multiply parentheses

7t^2-42t-(4(2t+3)-10)=0


We calculate terms in parentheses: -(4(2t+3)-10), so:

4(2t+3)-10

We multiply parentheses

8t+12-10

We add all the numbers together, and all the variables

8t+2

Back to the equation:

-(8t+2)


We get rid of parentheses

7t^2-42t-8t-2=0

We add all the numbers together, and all the variables

7t^2-50t-2=0

a = 7; b = -50; c = -2;
Δ = b2-4ac
Δ = -502-4·7·(-2)
Δ = 2556
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:

$\sqrt{\Delta}=\sqrt{2556}=\sqrt{36*71}=\sqrt{36}*\sqrt{71}=6\sqrt{71}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-50)-6\sqrt{71}}{2*7}=\frac{50-6\sqrt{71}}{14}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-50)+6\sqrt{71}}{2*7}=\frac{50+6\sqrt{71}}{14}
$