To solve the equation
\[ 6x^2 + x - 8 = 7, \]
let’s first move 7 to the left-hand side:
\[ 6x^2 + x - 8 - 7 = 0 \]
which simplifies to
\[ 6x^2 + x - 15 = 0. \]
Now, we will factor the quadratic expression \( 6x^2 + x - 15 \). We need to find two numbers that multiply to \( 6 \times (-15) = -90 \) and add to \( 1 \) (the coefficient of \( x \)).
The two numbers that meet these criteria are \( 10 \) and \( -9 \), since:
\[ 10 \times (-9) = -90 \quad \text{and} \quad 10 + (-9) = 1. \]
Next, we rewrite the equation by splitting the linear term using \( 10 \) and \( -9 \):
\[ 6x^2 + 10x - 9x - 15 = 0. \]
Now we can factor by grouping:
\[ (6x^2 + 10x) + (-9x - 15) = 0. \]
Factoring out \( 2x \) from the first group and \( -3 \) from the second group gives us:
\[ 2x(3x + 5) - 3(3x + 5) = 0. \]
Now, factor out \( 3x + 5 \):
\[ (3x + 5)(2x - 3) = 0. \]
Setting each factor to zero gives us the solutions:
-
\( 3x + 5 = 0 \)
- \( 3x = -5 \)
- \( x = -\frac{5}{3} \)
-
\( 2x - 3 = 0 \)
- \( 2x = 3 \)
- \( x = \frac{3}{2} \)
We already have one solution, which is \( \frac{3}{2} \). The other solution is
\[ -\frac{5}{3}. \]
Thus, the other solution is
\[ -\frac{5}{3}. \]
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