(d)=-d2+3d+4

Solution for (d)=-d2+3d+4 equation:

(d)=-d2+3d+4

We move all terms to the left:

(d)-(-d2+3d+4)=0

We add all the numbers together, and all the variables

-(-1d^2+3d+4)+d=0

We get rid of parentheses

1d^2-3d+d-4=0

We add all the numbers together, and all the variables

d^2-2d-4=0

a = 1; b = -2; c = -4;
Δ = b2-4ac
Δ = -22-4·1·(-4)
Δ = 20
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:

$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{5}}{2*1}=\frac{2-2\sqrt{5}}{2}
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{5}}{2*1}=\frac{2+2\sqrt{5}}{2}
$