Question

Solve the quadratic equation by factoring. 2x^2+5x+3=0

Answer 1

Answer:

$x_1=-\frac{3}{2} \\\\x_2=-1$

Step-by-step explanation:

the general fromula is

$(ax+b)(cx+)d)\\\\=acx^{2} +(ad+bc)x+bd$

so we need that

$ac=2\\\\ad+bc=5\\\\bd=3$

for ac = 2

we only have 2 and 1

so let's say

$a=2\\\\c=1$

so we now have

$(2x+b)(x+d)\\\\2x^{2} +(2d+b)x+bd$

and for bd = 3

we only have 3 and 1

but we need to

2d +b = 5

so we can have

b=3

and

d=1

so the factorized form is

$(2x+3)(x+1)$

so we come back to

$(2x+3)(x+1)=0$

so here we have two cases

Case 1

$(2x+3)=0\\\\2x+3=0\\\\2x=-3\\\\x=-\frac{3}{2}$

Case 2

$(x+1)=0\\\\x+1=0\\\\x=-1$

so those are our solutions

Answer 2

Answer:

-0.33 repeating

Step-by-step explanation:

2x^2=4x

4x+5x+3=0

combine like terms

9x + 3 = 0

subtract 3 from both sides

9x = -3

divide 9 both sides to get the variable by itself

x = -0.33 repeating