Question

Solve please 4x^2+2x-10=0

Answer 1

For this case we must solve a quadratic equation of the form $ax ^ 2 + bx + c = 0$, where its roots are given by:

$x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2a}$

If we have the following equation:

$4x ^ 2 + 2x-10 =0$

So:

$a = 4\\b = 2\\c = -10$

Substituting we have:

$x = \frac {-2 \pm \sqrt {2 ^ 2-4 (4) (- 10)}} {2 (4)}\\x = \frac {-2 \pm \sqrt {4 + 160}} {8}\\x = \frac {-2 \pm \sqrt {164}} {8}\\x = \frac {-2} {8} \pm \frac {\sqrt {164}} {8}\\x = \frac {-1} {4} \pm \frac {12.81} {8}$

So, we have:

$x_ {1} = - 0.25 + 1.60 = 1.35\\x_ {2} = - 0.25-1.60 = -1.85$

Answer:

The roots of the given equation are:

$x_ {1} = - 0.25 + 1.60 = 1.35\\x_ {2} = - 0.25-1.60 = -1.85$

Answer 2

Answer:

4x^2+2x=10

Step-by-step explanation:

you need to add the ten to both sides to get it away, and then you can't do any thing to the 4x^2 or the 2x