Question

3x² + 7x - 3 = 3x + 1

Explanation:

Step1: Rearrange the equation

Subtract \(3x\) and \(1\) from both sides to get the quadratic equation in standard form \(ax^2 + bx + c = 0\).
\(3x^2 + 7x - 3 - 3x - 1 = 0\)
Simplify the left - hand side: \(3x^2+(7x - 3x)+(-3 - 1)=0\), so \(3x^2 + 4x-4 = 0\)

Step2: Use the quadratic formula

For a quadratic equation \(ax^2+bx + c = 0\) (\(a
eq0\)), the quadratic formula is \(x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\).
In the equation \(3x^2 + 4x - 4 = 0\), we have \(a = 3\), \(b = 4\), and \(c=-4\).
First, calculate the discriminant \(\Delta=b^2 - 4ac=(4)^2-4\times3\times(-4)=16 + 48 = 64\)
Then, find the roots:
\(x=\frac{-4\pm\sqrt{64}}{2\times3}=\frac{-4\pm8}{6}\)

Step3: Calculate the two roots

For the plus sign: \(x_1=\frac{-4 + 8}{6}=\frac{4}{6}=\frac{2}{3}\)
For the minus sign: \(x_2=\frac{-4-8}{6}=\frac{-12}{6}=-2\)

Answer:

The solutions of the equation are \(x=\frac{2}{3}\) or \(x = - 2\)