Question

+ 4 = 20
-4 -4
x=1
x=-1
3 +3
20 = 0 -4
x² -4
4 20
(x-4)²=0
x-4=0
+4 +4
c. (x - 5)² =-
√(x-5)² = √
x-5 = √
x-5=±7
x-5=7 x-5
+5 +5 +
c. x² + 4x + 2
+20-20=0+
√(x+
solutions: x=
-1)³(-1)
b. 5x² + 5 =- 12x
a-5 b-5 c- -12

Explanation:

Step1: Rearrange the equation

First, we rearrange the given equation \(5x^{2}+5 = - 12x\) into standard quadratic form \(ax^{2}+bx + c=0\).
Add \(12x\) to both sides: \(5x^{2}+12x + 5=0\). Here, \(a = 5\), \(b = 12\), \(c = 5\).

Step2: Use the quadratic formula

The quadratic formula is \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\).
First, calculate the discriminant \(\Delta=b^{2}-4ac=(12)^{2}-4\times5\times5=144 - 100 = 44\).
Then, find the roots:
\(x=\frac{-12\pm\sqrt{44}}{2\times5}=\frac{-12\pm2\sqrt{11}}{10}=\frac{-6\pm\sqrt{11}}{5}\)

Answer:

\(x=\frac{-6 + \sqrt{11}}{5}\) or \(x=\frac{-6-\sqrt{11}}{5}\) (or approximately \(x\approx\frac{-6 + 3.3166}{5}\approx - 0.5367\) and \(x\approx\frac{-6-3.3166}{5}\approx - 1.8633\))