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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

(4d+5)(2d+3)=0

We multiply parentheses ..

(+8d^2+12d+10d+15)=0

We get rid of parentheses

8d^2+12d+10d+15=0

We add all the numbers together, and all the variables

8d^2+22d+15=0

a = 8; b = 22; c = +15;
Δ = b2-4ac
Δ = 222-4·8·15
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-2}{2*8}=\frac{-24}{16}
=-1+1/2
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+2}{2*8}=\frac{-20}{16}
=-1+1/4
$