Respuesta :

You use the quadratic formula:
[tex]2x {}^{2} - 3x = 5[/tex]
[tex]2x {}^{2} - 3x - 5 = 0[/tex]
[tex]x = \frac{3 + - \sqrt{9 -4(2)( - 5)}}{2 \times 2} [/tex]
[tex]x = \frac{3 + - \sqrt{49} }{4} [/tex]
[tex]x = \frac{3 + 7}{4} \: and \: x = \frac{3 - 7}{4} [/tex]
[tex]x = \frac{5}{2} \: and \: x = - 1[/tex]
Ver imagen demariaa01

Answer:

Given the equation:

[tex]2x^2-3x=5[/tex]

Subtract 5 from both sides we get;

[tex]2x^2-3x-5 = 0[/tex]

Factorize the equation:

Break the middle term we get;

[tex]2x^2-5x+2x-5 = 0[/tex]

⇒[tex]x(2x-5)+1(2x-5)=0[/tex]

Take (2x-5) common we have;

[tex](2x-5)(x+1) = 0[/tex]

By zero product property we have;

[tex]2x-5 = 0[/tex]   or [tex]x+1 = 0[/tex]

⇒[tex]x = \frac{5}{2}[/tex] or x = -1

Therefore, the solutions for the given equation are:

[tex]\frac{5}{2}[/tex], [tex]-1[/tex]

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