solve the equation 3m^2+6m=3 using the quadratic formula?

Answer 1

U Serious?
OK
The formula is:
[-b(+/-)√(b²-4ac)]/2a
for:
ax²+bx+c=0

So:
3m²+6m-3=0

and we substitute:

[-6(+/)√(6²-4(3)(-3))]/2(3)
[-6(+/-)√(36+36)]/6
(-6(+/-)√72)/6

Since we can factorize 72, we get √72=√2*2*2*3*3
√4*√2√9=2*3√2=6√2

So we get the two answers:
(-6+6√2)/6 and (-6-6√2)/6
6(-1+√2)/6 and 6(-1-√2)/6
or:
-1+√2 and -1-√2

Answer 2

3m^2+6m=3
=>3m(m+ 2) = 3
=>m(m+2) = 3/3
=>m(m+2) = 1
=>m^2 + 2m = 1
=>m^2 + 2m - 1 = 0

a = 1, b = 2, c = -1

m = -b + or - sqrt (b^2 - 4ac) / 2a
m = - 2 + - sqrt { (2)^2 - 4 * 1 * -1 } / 2 * 1
m = - 2 + - sqrt ( 4 + 4 ) / 2
m = - 2 + - sqrt ( 8) ./ 2
m = -2 + - sqrt ( 2 *4) / 2
m = - 2 + - 2 sqrt (2) / 2
so
m = -2 + 2 sqrt ( 2) / 2 or - 2 - 2sqrt ( 2) / 2
m = 2( - 1 + sqrt ( 2) / 2 or -2( 1 + sqrt(2) / 2
m = sqrt (2) - 1 or -sqrt(2) - 1

Answer 3

3m^2 + 6m - 3 = 0 set equal to zero first always

m = -1 +- sqrt (2)

Answer 4

3m^2 + 6m - 3 = 0

a = 3 b = 6 c = -3

m = [-b +- sqrt(b^2 - 4ac) ] / 2a
m = [-6 +- sqrt(36 - 4(3)(-3)) ] / 6
m = [-6 +- sqrt(36 + 36) ] / 6
m = [-6 +- sqrt(72) ] / 6
m = [-6 +- 6sqrt(2) ] / 6
m = -1 +- sqrt(2)
m = .414 or -2.414 (approximately)

Answer 5

3m² + 6m - 3 = 0
m = [- 6 ± √(36 + 36) ] / 6
m = [- 6 ± √ 72 ] / 6
m = [- 6 ± 6√2 ] / 6
m = -1 ± √2
m = 0.414 , m = - 2.414

Answer 6

9

Answer 7

3m^2+6m-3=0

divide through by 3 to get:

m^2+2m-1=0

so m = -2 plus/minus sqrt(4-4*1*(-1)) / 2

m = -2 +/- sqrt(8) / 2

m = -1 +/- sqrt(8)/2

m = -1 +/- sqrt(4)sqrt(2)/2

m = -1 +/- 2sqrt(2)/2

m = -1 +/- sqrt(2)

Answer 8

3m^2+6m=3
3(m^2 + 2m - 1) = 0
m = ( -2 +/- sqrt(4 + 4) ) / 2
= ( - 2 +/- 2sqrt(2) ) / 2
= -1 +/- sqrt(2).

Answer 9

no clue sorry