Question

Please solve the following equation.​

Answer 1

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Answer 2

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$2x^2 -3x =20$

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Given:

$2x^2-3x =20$

$2x^2-3x-20=0$

Now:

• The product of middle terms should be $-20 × 2 = -40$
• The sum of middle terms should be $-3$

Making the Factors:

$2x^2-8x+5x-20=0$

$2x(x-4)+5(x-4)=0$

$(x-4)(2x+5)=0$

Thus:

$x= \displaystyle\frac{-5}{2}$

$x= 4$

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