5(3t^2+2t+4)=5(5t+8)

Solution for 5(3t^2+2t+4)=5(5t+8) equation:

5(3t^2+2t+4)=5(5t+8)

We move all terms to the left:

5(3t^2+2t+4)-(5(5t+8))=0

We multiply parentheses

15t^2+10t-(5(5t+8))+20=0


We calculate terms in parentheses: -(5(5t+8)), so:

5(5t+8)

We multiply parentheses

25t+40

Back to the equation:

-(25t+40)


We get rid of parentheses

15t^2+10t-25t-40+20=0

We add all the numbers together, and all the variables

15t^2-15t-20=0

a = 15; b = -15; c = -20;
Δ = b2-4ac
Δ = -152-4·15·(-20)
Δ = 1425
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:

$\sqrt{\Delta}=\sqrt{1425}=\sqrt{25*57}=\sqrt{25}*\sqrt{57}=5\sqrt{57}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-5\sqrt{57}}{2*15}=\frac{15-5\sqrt{57}}{30}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+5\sqrt{57}}{2*15}=\frac{15+5\sqrt{57}}{30}
$