(3x+50)(7x-10)=15

Solution for (3x+50)(7x-10)=15 equation:

(3x+50)(7x-10)=15

We move all terms to the left:

(3x+50)(7x-10)-(15)=0

We multiply parentheses ..

(+21x^2-30x+350x-500)-15=0

We get rid of parentheses

21x^2-30x+350x-500-15=0

We add all the numbers together, and all the variables

21x^2+320x-515=0

a = 21; b = 320; c = -515;
Δ = b2-4ac
Δ = 3202-4·21·(-515)
Δ = 145660
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:

$\sqrt{\Delta}=\sqrt{145660}=\sqrt{4*36415}=\sqrt{4}*\sqrt{36415}=2\sqrt{36415}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(320)-2\sqrt{36415}}{2*21}=\frac{-320-2\sqrt{36415}}{42}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(320)+2\sqrt{36415}}{2*21}=\frac{-320+2\sqrt{36415}}{42}
$