1445=4x^2+19x

Solution for 1445=4x^2+19x equation:

1445=4x^2+19x

We move all terms to the left:

1445-(4x^2+19x)=0

We get rid of parentheses

-4x^2-19x+1445=0

a = -4; b = -19; c = +1445;
Δ = b2-4ac
Δ = -192-4·(-4)·1445
Δ = 23481
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{23481}=\sqrt{9*2609}=\sqrt{9}*\sqrt{2609}=3\sqrt{2609}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-3\sqrt{2609}}{2*-4}=\frac{19-3\sqrt{2609}}{-8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+3\sqrt{2609}}{2*-4}=\frac{19+3\sqrt{2609}}{-8}
$