Question

21x² + 26x - 5 = 0

Explanation:

The given equation seems to be \(21x^{2}+26x - 15 = 0\) (assuming the last part is \(-15 = 0\) from the hand - written text). We can solve this quadratic equation using the quadratic formula. For a quadratic equation \(ax^{2}+bx + c = 0\), the solutions are given by \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\).

Step 1: Identify the coefficients

For the equation \(21x^{2}+26x - 15 = 0\), we have \(a = 21\), \(b = 26\) and \(c=- 15\).

Step 2: Calculate the discriminant \(\Delta=b^{2}-4ac\)

Substitute the values of \(a\), \(b\) and \(c\) into the discriminant formula:
\(\Delta=(26)^{2}-4\times21\times(-15)\)
\(=676 + 1260\)
\(=1936\)

Step 3: Find the square root of the discriminant

\(\sqrt{\Delta}=\sqrt{1936} = 44\)

Step 4: Calculate the roots using the quadratic formula

\(x=\frac{-b\pm\sqrt{\Delta}}{2a}=\frac{-26\pm44}{2\times21}\)

We have two cases:

• Case 1: When we take the plus sign

\(x=\frac{-26 + 44}{42}=\frac{18}{42}=\frac{3}{7}\)

• Case 2: When we take the minus sign

\(x=\frac{-26-44}{42}=\frac{-70}{42}=-\frac{5}{3}\)

The solutions of the quadratic equation \(21x^{2}+26x - 15 = 0\) are \(x=\frac{3}{7}\) and \(x =-\frac{5}{3}\)

Answer:

The given equation seems to be \(21x^{2}+26x - 15 = 0\) (assuming the last part is \(-15 = 0\) from the hand - written text). We can solve this quadratic equation using the quadratic formula. For a quadratic equation \(ax^{2}+bx + c = 0\), the solutions are given by \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\).

Step 1: Identify the coefficients

For the equation \(21x^{2}+26x - 15 = 0\), we have \(a = 21\), \(b = 26\) and \(c=- 15\).

Step 2: Calculate the discriminant \(\Delta=b^{2}-4ac\)

Substitute the values of \(a\), \(b\) and \(c\) into the discriminant formula:
\(\Delta=(26)^{2}-4\times21\times(-15)\)
\(=676 + 1260\)
\(=1936\)

Step 3: Find the square root of the discriminant

\(\sqrt{\Delta}=\sqrt{1936} = 44\)

Step 4: Calculate the roots using the quadratic formula

\(x=\frac{-b\pm\sqrt{\Delta}}{2a}=\frac{-26\pm44}{2\times21}\)

We have two cases:

• Case 1: When we take the plus sign

\(x=\frac{-26 + 44}{42}=\frac{18}{42}=\frac{3}{7}\)

• Case 2: When we take the minus sign

\(x=\frac{-26-44}{42}=\frac{-70}{42}=-\frac{5}{3}\)

The solutions of the quadratic equation \(21x^{2}+26x - 15 = 0\) are \(x=\frac{3}{7}\) and \(x =-\frac{5}{3}\)