Sure, let's solve the equation [tex]\( 4w^2 = 15w + 4 \)[/tex].
First, we set the equation to zero by bringing all terms to one side:
[tex]\[ 4w^2 - 15w - 4 = 0 \][/tex]
This is a quadratic equation in the standard form [tex]\( ax^2 + bx + c = 0 \)[/tex].
Here, [tex]\( a = 4 \)[/tex], [tex]\( b = -15 \)[/tex], and [tex]\( c = -4 \)[/tex].
To solve the quadratic equation, we use the quadratic formula:
[tex]\[ w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Substitute [tex]\( a = 4 \)[/tex], [tex]\( b = -15 \)[/tex], and [tex]\( c = -4 \)[/tex] into the formula:
[tex]\[ w = \frac{-(-15) \pm \sqrt{(-15)^2 - 4 \cdot 4 \cdot (-4)}}{2 \cdot 4} \][/tex]
[tex]\[ w = \frac{15 \pm \sqrt{225 + 64}}{8} \][/tex]
[tex]\[ w = \frac{15 \pm \sqrt{289}}{8} \][/tex]
[tex]\[ w = \frac{15 \pm 17}{8} \][/tex]
Now, we solve for the two possible values of [tex]\( w \)[/tex]:
For the positive case:
[tex]\[ w = \frac{15 + 17}{8} \][/tex]
[tex]\[ w = \frac{32}{8} \][/tex]
[tex]\[ w = 4 \][/tex]
For the negative case:
[tex]\[ w = \frac{15 - 17}{8} \][/tex]
[tex]\[ w = \frac{-2}{8} \][/tex]
[tex]\[ w = -\frac{1}{4} \][/tex]
So, the solutions to the equation [tex]\( 4w^2 = 15w + 4 \)[/tex] are:
[tex]\[ w = 4, -\frac{1}{4} \][/tex]
