Solve for: 2x^2-11x+15=0

Expression: $2{x}^{2}-11x+15=0$

Identify the coefficients $a$, $b$ and $c$ of the quadratic equation

$\begin{array} { l }a=2,& b=-11,& c=15\end{array}$

Substitute $a=2$, $b=-11$ and $c=15$ into the quadratic formula $x=\frac{ -b\pm\sqrt{ {b}^{2}-4ac } }{ 2a }$

$x=\frac{ -\left( -11 \right)\pm\sqrt{ {\left( -11 \right)}^{2}-4 \times 2 \times 15 } }{ 2 \times 2 }$

When there is a $-$ in front of an expression in parentheses, change the sign of each term of the expression and remove the parentheses

$x=\frac{ 11\pm\sqrt{ {\left( -11 \right)}^{2}-4 \times 2 \times 15 } }{ 2 \times 2 }$

Evaluate the power

$x=\frac{ 11\pm\sqrt{ 121-4 \times 2 \times 15 } }{ 2 \times 2 }$

Calculate the product

$x=\frac{ 11\pm\sqrt{ 121-120 } }{ 2 \times 2 }$

Multiply the numbers

$x=\frac{ 11\pm\sqrt{ 121-120 } }{ 4 }$

Subtract the numbers

$x=\frac{ 11\pm\sqrt{ 1 } }{ 4 }$

Any root of $1$ equals $1$

$x=\frac{ 11\pm1 }{ 4 }$

Write the solutions, one with a $+$ sign and one with a $-$ sign

$\begin{array} { l }x=\frac{ 11+1 }{ 4 },\\x=\frac{ 11-1 }{ 4 }\end{array}$

Simplify the expression

$\begin{array} { l }x=3,\\x=\frac{ 11-1 }{ 4 }\end{array}$

Simplify the expression

$\begin{array} { l }x=3,\\x=\frac{ 5 }{ 2 }\end{array}$

The equation has $2$ solutions

$\begin{align*}&\begin{array} { l }x_1=\frac{ 5 }{ 2 },& x_2=3\end{array} \\&\begin{array} { l }x_1=2.5,& x_2=3\end{array}\end{align*}$