Question

Solve using quadratic formula; Leave irrational answers in radical form:
x^2+2x-3=0

Answer 1

noneed
x^(2)+2x-3=0

(x+3)(x-1)=0

x+3=0_x-1=0

x=-3_x-1=0

x=-3_x-1=0

x=-3_x=1

x=-3,1

Answer 2

If you were to use the quadratic equation to solve something like this \[0 = ax^2 + bx + c\]substitute your coefficients into this and solve.\[x = \frac {-b \pm \sqrt {b^2 - 4ac}}{2a}\]

Answer 3

i dont understand :/

Answer 4

Alright, your equation is
x^2+2x-3=0
so your coefficients are
a = 1, b = 2, c = -3

Now plug them into the quadratic formula to sovle for x.

Answer 5

can you work the problem out on here & show me.

Answer 6

quadratic formula is applicable in finding solution x if the given quadratic equation is non-factorable. .

your given quadratic equation is factorable, that's why there is no need for using quadratic formula. .

Answer 7

\[x = \frac {-b \pm \sqrt {b^2 - 4ac}}{2a} = \frac {-2 \pm \sqrt {2^2 - 4 \times 1 \times -3}}{2 \times 1} = \frac {-2 \pm 4}{2}\]\[x = -1 \pm 2 \]For +, x = 1, for -, x = -3

Answer 8

Yeah, as Sam and jerwyn said, its not necessary to use the quadratic equation on something like since its factorable, but since the question wants you to do it this way, you have no choice =P