Hi there!
The question gives us the quadratic equation , and it tells us to solve it using the quadratic formula, which goes as . However, we must first find the values of a, b, and c. The official quadratic equation goes as , which matches the format of the given quadratic equation. Hence, the value of a would be 1, the value of b would be 5, and the value of c would be 3. Now, just plug it back into the quadratic equation and simplify to get the zeros of the equation.
x = \frac{-b \pm \sqrt{b^2 - 4ac} }{2a}
x = \frac{-(5) \pm \sqrt{(5)^2 - 4(1)(3)} }{2(1)}
x = \frac{-5 \pm \sqrt{25 - 12} }{2}
x = \frac{-5 \pm \sqrt{13} }{2}
x = \frac{-5 \pm 3.61 }{2}
x = \frac{-5 + 3.61 }{2}, x = \frac{-5 - 3.61 }{2}
x=-0.695 \ \textgreater \ \ \textgreater \ -0.7, x= -4.305 \ \textgreater \ \ \textgreater \ x=-4.31
Therefore, the solutions to the quadratic equation are x = -0.7 and x = -4.31. Hope this helped and have a phenomenal day!
Your answer is 4.31
Answer:
-1/4x
Step-by-step explanation:
11-12/6-2 = -1/4
i believe it would be: 1/2y = x
y = 2.5
Answer:
x=7, x= -2
Step-by-step explanation:
- Start by putting the whole equation to one side: x2-5x=14 ----> x2-5x-14=0
- Next, factor the equation: x,x -7,2 ----> (x-7)(x+2)=0
- Now, solve both equations: x-7=0....x=7 AND x+2=0.....x= -2
- Plug both in to make sure that they are correct....both are correct
