How do you solve #4x^2 +4x = 15# using the quadratic formula?

First Of All, Convert the Equation to It's General Form #ax^2 + bx + c = 0#.

So We have,

#color(white)(xxx)4x^2 + 4x = 15#

#rArr 4x^2 + 4x - 15 = 0# [Subtract #15# from both sides.]

So, Comparing the Equation with the General Form, We get,

#a = 4, b = 4, c = -15#.

So, Let's Find the Discriminat.

#D = b^2 - 4ac = 4^2 - 4*4*(-15) = 16 + 240 = 256#

As #D gt 0#, we will get two roots which are real and distinct.

Now Use The Quadratic Formula or Sridhar Acharya's Rule (whatever you may call it in your country).

#alpha = (-b + sqrt(D))/(2a) = (-4 + sqrt(256))/(2 * 4) = (-4 + 16)/(8) = 3/2#

And #beta = (-b - sqrt(D))/(2a) = (-4 - sqrt(256))/(2 * 4) = (-4 - 16)/(8) = 5/2#

So, #x = 3/2, 5/2#

Hope this helps.

Make the expression equal to zero:

#4 x^2 + 4x - 15 = 0#

The quadratic formula is:

# x = (-b +- sqrt (b^2 -4 a c )) / (2a) #

In our case we substitute:

# a = 4, b = 4, c = -15#

So that the quadratic formula becomes:

# x = (-4 +- sqrt (4^2 -4 * 4 *(-15) )) / (2*4) #

# = (-4 +- sqrt (16 + 240 )) / (8) #

# = (-4 +- 16) / (8) #

# = 3/2 "or" - 5/2 #

We can also solve this by factorising (here you need to guess)

#4 x^2 + 4x - 15 = 0#

#(2x + 5 ) (2x - 3) = 0 #

Then make both parenthesis equal to zero to get the same answers:

#(2x + 5 ) = 0 " or " (2x - 3) = 0 #