How do you solve $2x²+3x=5$ using the quadratic formula?

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Answer 1 · 2017-02-11T02:28:56

$- 5/2, 1$

Bring the equation to standard form:

$y = 2x^2 + 3x - 5 = 0$ .

$D = d^2 = b^2 - 4ac = 9 + 40 = 49$ --> $d = +- 7$

There are 2 real roots:

$x = - b/(2a) +- d/(2a) = - 3/4 +- 7 /4$

$x1 =( - 3 - 7)/4 = - 10/4 = -5/2$

$x2 = (- 3 + 7)/4 = 4/4 = 1$

How do you solve 2x²+3x=5 2x²+3x=5 using the quadratic formula?