((2x-3)(x+5))/2=240

Solution for ((2x-3)(x+5))/2=240 equation:

((2x-3)(x+5))/2=240

We move all terms to the left:

((2x-3)(x+5))/2-(240)=0

We multiply parentheses ..

((+2x^2+10x-3x-15))/2-240=0

We multiply all the terms by the denominator


((+2x^2+10x-3x-15))-240*2=0


We calculate terms in parentheses: +((+2x^2+10x-3x-15)), so:

(+2x^2+10x-3x-15)

We get rid of parentheses

2x^2+10x-3x-15

We add all the numbers together, and all the variables

2x^2+7x-15

Back to the equation:

+(2x^2+7x-15)


We add all the numbers together, and all the variables

(2x^2+7x-15)-480=0

We get rid of parentheses

2x^2+7x-15-480=0

We add all the numbers together, and all the variables

2x^2+7x-495=0

a = 2; b = 7; c = -495;
Δ = b2-4ac
Δ = 72-4·2·(-495)
Δ = 4009
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-\sqrt{4009}}{2*2}=\frac{-7-\sqrt{4009}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{4009}}{2*2}=\frac{-7+\sqrt{4009}}{4}
$