please solve this quadratic formula?

Factoring

(3x+ 7)(3x-3)/3=0
(3x+7)(x-1)=0
x=-7/3 and x=1

Hope this help

DG

3x² + 4x - 7 = 0
x² + 4/3x = 7/3
x² + 2/3x = 7/3 + (2/3)²
x² + 2/3x = 21/9 + 4/9
(x + 2/3)² = 25/9
x + 2/3 = +/- 5/3

x = 5/3 - 2/3, x = 3/3, x = 1
x = - 5/3 - 2/3, x = - 7/3

Answer: x = 1, - 7/3

Proof (x = 1):
3(1)² + 4(1) = 7
3(1) + 4 = 7
3 + 4 = 7

Proof (x = - 7/3):
3(- 7/3)² + 4(- 7/3) = 7
3(49/9) - 28/3 = 7
49/3 - 28/3 = 7
21/3 = 7
7 = 7

This is not a quadratic formula. This is a quadratic equation.

3x^2 + 4x - 7 = 0
3x^2 + 7x - 3x - 7 = 0
(3x^2 + 7x) - (3x + 7) = 0
x(3x + 7) - 1(3x + 7) = 0
(3x + 7)(x - 1) = 0

3x + 7 = 0
3x = -7
x = -7/3

x - 1 = 0
x = 1

∴ x = -7/3, 1

3x^2+4x -7=0
we can solve this equation by factorization
we have to split 4x into 2 terms(let's say d1 and d2) such that
d1*d2=a*c=3x^2*(-7)=-21x^2
and d1+d2=4x
such 2 terms are 7x and (-3)x
so 3x^2+4x-7=0
3x^2+(-3)x+7x-7=0
3x(x-1)+7(x-1)=0 (taking 3x and 7 common)
(3x+7)(x-1)=0 ( as a(b)+c(b)=(a+c)b
so either 3x+7=0 which means x=-7/3
or x-1=0 which means x=1

3x^2 + 4x - 7 = 0

x(3x + 4) - 7 = 0

(3x + 4) (x - 7) = 0

x = -4/3 or x = 7

x = [ - b ± √ ( b ² - 4 a c ) ] / 2 a

x = [ - 4 ± √ ( 16 + 84 ) ] / 6

x = [ - 4 ± √ ( 100 ) ] / 6

x = [ - 4 ± 10 ] / 6

x = 6/6 , x = - 14/6

x = 1 , x = - 7/3

OR

( 3 x + 7 ) ( x - 1 ) = 0

x = - 7/3 , x = 1

A trick when factoring quadratics that have a leading coefficient greater than 1 is to do this...

Multiply 3 x (-7)

3 x -7 = -21

Find the factors of -21 that will add to the middle term (+4)

7 and -3

Now rewrite the equatioin, but substitute 7x and -3x for the middle terms

3x^2 - 3x + 7x - 7 = 0

factor by grouping

3x (x - 1) + 7 (x - 1) = 0

(3x + 7) (x - 1) = 0

Now solve each for x

3x + 7 = 0
x = -7/3

x - 1 = 0
x = 1

the solution set is {-7/3, 1}

(3x +7)(x-1 )
x= -7/3, 1

3x^2+4x-7=0

a=3
b=4 c=-7

-4+/- sq rt 4^2-4(3)(-7)/6

-4 +/- sq rt 16+84/6
-4 +/- sq rt 100/6

-4+/- 10/6
x=-2 +/- 5/3
x= -11/3 or -1/3