(3+x)(2x-5)=78

Solution for (3+x)(2x-5)=78 equation:

(3+x)(2x-5)=78

We move all terms to the left:

(3+x)(2x-5)-(78)=0

We add all the numbers together, and all the variables

(x+3)(2x-5)-78=0

We multiply parentheses ..

(+2x^2-5x+6x-15)-78=0

We get rid of parentheses

2x^2-5x+6x-15-78=0

We add all the numbers together, and all the variables

2x^2+x-93=0

a = 2; b = 1; c = -93;
Δ = b2-4ac
Δ = 12-4·2·(-93)
Δ = 745
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{-(1)-\sqrt{745}}{2*2}=\frac{-1-\sqrt{745}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{745}}{2*2}=\frac{-1+\sqrt{745}}{4}
$