Question

solve (2x) (3x-7) [(2/3x -8] =0

Answer 1

Answer:

To solve using the quadratic⬆️

2x2−3x−7=0

Step-by-step explanation:

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Answer 2

$\sf\huge\bold{\underline{\underline{{Solution}}}}$

$\sf \longmapsto(2x)(3x - 7) \bigg( \dfrac{2}{3} x - 8 \bigg) = 0$

How to solve:

Step1:first step is to solve terms inside brackets and then Multiply each and every term with outside term.

Step2:Arrange the term so that like terms gets together.

Step3: Evaluate it and simply it in the order of degree of polynomial if terms comes in quadratic then factorise it or let leave alone .

$\sf \longmapsto6 {x}^{2} - 14x \bigg( \dfrac{2x - 24}{3} \bigg) = 0$

$\sf \longmapsto \dfrac{(6 {x}^{2} - 14x)(2x - 24)}{3} = 0$

Cross multiply to both sides

$\sf \longmapsto(6 {x}^{2} - 14x)(2x - 24) = 3 \times 0$

$\sf \longmapsto2x(6 {x}^{2} - 14x) - 24(6 {x}^{2} - 14x) = 0$

$\sf \longmapsto12 {x}^{3} - 28 {x}^{2} - 14 {x}^{2} + 336 = 0$

$\sf \longmapsto12 {x}^{3} - 172 {x}^{2} + 336 = 0$

Taking 4 common

$\sf \longmapsto4(3 {x}^{3} - 43x + 84) = 0$

$\sf = 3 {x}^{3} - 43 {x}^{2} + 84$