Question

Решите уравнение, прошу: (7х+1)(х-3)+20(х-1)(х+1)=3(3х-2)²+13

Answer 1

7х²-21х+х-3+20х²-20=27х²-24х+25
27х²-20х-23=27х²-24х+25
4х=48
х=12

Answer 2

8х+20х+1х_1+3+1_1_9+13 вот

Answer 3

I can help you solve this equation. First, let's expand the brackets and simplify the terms:

$$ (7x+1)(x-3)+20(x-1)(x+1)=3(3x-2)^2+13 $$

$$ 7x^2-20x-3+20x^2-20x+20=27x^2-36x+12+13 $$

$$ 4x^2-36x+2=0 $$

Next, let's use the quadratic formula to find the values of x that satisfy the equation:

$$ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} $$

In this case, $a=4$, $b=-36$, and $c=2$. Plugging these values into the formula, we get:

$$ x=\frac{36\pm\sqrt{1296-32}}{8} $$

$$ x=\frac{36\pm\sqrt{1264}}{8} $$

$$ x=\frac{36\pm 4\sqrt{79}}{8} $$

Finally, let's simplify the fractions by dividing both the numerator and denominator by 4:

$$ x=\frac{9\pm\sqrt{79}}{2} $$

These are the two solutions for x. I hope this helps you understand how to solve the equation.Ответь+подробно.+Решите+уравнение,+прошу:+(7х+1)(х-3)+20(х-1)(х+1)=3(3х-2)²+13