(21x+5)(10x-3)=147

Solution for (21x+5)(10x-3)=147 equation:

(21x+5)(10x-3)=147

We move all terms to the left:

(21x+5)(10x-3)-(147)=0

We multiply parentheses ..

(+210x^2-63x+50x-15)-147=0

We get rid of parentheses

210x^2-63x+50x-15-147=0

We add all the numbers together, and all the variables

210x^2-13x-162=0

a = 210; b = -13; c = -162;
Δ = b2-4ac
Δ = -132-4·210·(-162)
Δ = 136249
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{-(-13)-\sqrt{136249}}{2*210}=\frac{13-\sqrt{136249}}{420}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+\sqrt{136249}}{2*210}=\frac{13+\sqrt{136249}}{420}
$