-4d^2+9d+6=3-7d

Solution for -4d^2+9d+6=3-7d equation:

-4d^2+9d+6=3-7d

We move all terms to the left:

-4d^2+9d+6-(3-7d)=0

We add all the numbers together, and all the variables

-4d^2+9d-(-7d+3)+6=0

We get rid of parentheses

-4d^2+9d+7d-3+6=0

We add all the numbers together, and all the variables

-4d^2+16d+3=0

a = -4; b = 16; c = +3;
Δ = b2-4ac
Δ = 162-4·(-4)·3
Δ = 304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{304}=\sqrt{16*19}=\sqrt{16}*\sqrt{19}=4\sqrt{19}$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4\sqrt{19}}{2*-4}=\frac{-16-4\sqrt{19}}{-8}
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4\sqrt{19}}{2*-4}=\frac{-16+4\sqrt{19}}{-8}
$