4(d+7);d=3

Solution for 4(d+7);d=3 equation:

4(d+7)d=3

We move all terms to the left:

4(d+7)d-(3)=0

We multiply parentheses

4d^2+28d-3=0

a = 4; b = 28; c = -3;
Δ = b2-4ac
Δ = 282-4·4·(-3)
Δ = 832
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:

$\sqrt{\Delta}=\sqrt{832}=\sqrt{64*13}=\sqrt{64}*\sqrt{13}=8\sqrt{13}$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-8\sqrt{13}}{2*4}=\frac{-28-8\sqrt{13}}{8}
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+8\sqrt{13}}{2*4}=\frac{-28+8\sqrt{13}}{8}
$