3d2+4=10;d

Solution for 3d2+4=10;d equation:

3d^2+4=10d

We move all terms to the left:

3d^2+4-(10d)=0

a = 3; b = -10; c = +4;
Δ = b2-4ac
Δ = -102-4·3·4
Δ = 52
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:

$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{13}}{2*3}=\frac{10-2\sqrt{13}}{6}
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{13}}{2*3}=\frac{10+2\sqrt{13}}{6}
$