7d2+10d=3

Solution for 7d2+10d=3 equation:

7d^2+10d=3

We move all terms to the left:

7d^2+10d-(3)=0

a = 7; b = 10; c = -3;
Δ = b2-4ac
Δ = 102-4·7·(-3)
Δ = 184
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:

$\sqrt{\Delta}=\sqrt{184}=\sqrt{4*46}=\sqrt{4}*\sqrt{46}=2\sqrt{46}$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{46}}{2*7}=\frac{-10-2\sqrt{46}}{14}
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{46}}{2*7}=\frac{-10+2\sqrt{46}}{14}
$