(2x+5)(4x+19)=18-

Solution for (2x+5)(4x+19)=18- equation:

(2x+5)(4x+19)=18-

We move all terms to the left:

(2x+5)(4x+19)-(18-)=0

We add all the numbers together, and all the variables

(2x+5)(4x+19)-0=0

We add all the numbers together, and all the variables

(2x+5)(4x+19)=0

We multiply parentheses ..

(+8x^2+38x+20x+95)=0

We get rid of parentheses

8x^2+38x+20x+95=0

We add all the numbers together, and all the variables

8x^2+58x+95=0

a = 8; b = 58; c = +95;
Δ = b2-4ac
Δ = 582-4·8·95
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(58)-18}{2*8}=\frac{-76}{16}
=-4+3/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(58)+18}{2*8}=\frac{-40}{16}
=-2+1/2
$