5(3d+2)+15=4(2d+5)7d+5

Solution for 5(3d+2)+15=4(2d+5)7d+5 equation:

5(3d+2)+15=4(2d+5)7d+5

We move all terms to the left:

5(3d+2)+15-(4(2d+5)7d+5)=0

We multiply parentheses

15d-(4(2d+5)7d+5)+10+15=0


We calculate terms in parentheses: -(4(2d+5)7d+5), so:

4(2d+5)7d+5

We multiply parentheses

56d^2+140d+5

Back to the equation:

-(56d^2+140d+5)


We add all the numbers together, and all the variables

15d-(56d^2+140d+5)+25=0

We get rid of parentheses

-56d^2+15d-140d-5+25=0

We add all the numbers together, and all the variables

-56d^2-125d+20=0

a = -56; b = -125; c = +20;
Δ = b2-4ac
Δ = -1252-4·(-56)·20
Δ = 20105
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-125)-\sqrt{20105}}{2*-56}=\frac{125-\sqrt{20105}}{-112}
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-125)+\sqrt{20105}}{2*-56}=\frac{125+\sqrt{20105}}{-112}
$