What is the answer to (3x+2)(7x-5)=21x^2-x-10 ?

The answer to (3x+2)(7x-5)=21x^2-x-10 is True for all x

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(3x+2)(7x-5)=21x^2-x-10

(3 x + 2) (7 x - 5) = 21 x^{2} - x - 10

True for all x

Solution steps

(3 x + 2) (7 x - 5) = 21 x^{2} - x - 10

Expand

(3 x + 2) (7 x - 5) : 21 x^{2} - x - 10

21 x^{2} - x - 10 = 21 x^{2} - x - 10

Move

10

to the right side

21 x^{2} - x = 21 x^{2} - x

Subtract

21 x^{2} - x

from both sides

21 x^{2} - x - (21 x^{2} - x) = 21 x^{2} - x - (21 x^{2} - x)

Simplify

0 = 0

Both sides are equal

True for all x

\frac{1}{2} lo g_{5} (15) - lo g_{5} (75)

7 \cdot \frac{1}{5}

2 x + 3 (1 - 2 x) < x + 8

212.9 \cdot (\frac{39.3701}{1}) \cdot (\frac{60}{1})

x - 1 \geq 5

What is the answer to (3x+2)(7x-5)=21x^2-x-10 ?
The answer to (3x+2)(7x-5)=21x^2-x-10 is True for all x
Solve for x: (3x+2)(7x-5)=21x^2-x-10
The solution is True for all x