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

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

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

We move all terms to the left:

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

We multiply parentheses

7t^2+28t-(4(3t+2))-10=0


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

4(3t+2)

We multiply parentheses

12t+8

Back to the equation:

-(12t+8)


We get rid of parentheses

7t^2+28t-12t-8-10=0

We add all the numbers together, and all the variables

7t^2+16t-18=0

a = 7; b = 16; c = -18;
Δ = b2-4ac
Δ = 162-4·7·(-18)
Δ = 760
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:

$\sqrt{\Delta}=\sqrt{760}=\sqrt{4*190}=\sqrt{4}*\sqrt{190}=2\sqrt{190}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{190}}{2*7}=\frac{-16-2\sqrt{190}}{14}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{190}}{2*7}=\frac{-16+2\sqrt{190}}{14}
$