Question

What is the solution set of 2x 2 - 3x = 5?

Answer 1

Answer:

Given the equation:

$2x^2-3x=5$

Subtract 5 from both sides we get;

$2x^2-3x-5 = 0$

Factorize the equation:

Break the middle term we get;

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

⇒$x(2x-5)+1(2x-5)=0$

Take (2x-5) common we have;

$(2x-5)(x+1) = 0$

By zero product property we have;

$2x-5 = 0$   or $x+1 = 0$

⇒$x = \frac{5}{2}$ or x = -1

Therefore, the solutions for the given equation are:

$\frac{5}{2}$, $-1$

Answer 2

You use the quadratic formula:
$2x {}^{2} - 3x = 5$
$2x {}^{2} - 3x - 5 = 0$
$x = \frac{3 + - \sqrt{9 -4(2)( - 5)}}{2 \times 2}$
$x = \frac{3 + - \sqrt{49} }{4}$
$x = \frac{3 + 7}{4} \: and \: x = \frac{3 - 7}{4}$
$x = \frac{5}{2} \: and \: x = - 1$