Solve $2x^2+4x+2=x^24x13$2x2+4x+2=x2−4x−13 for $x$x.
Think: Since there are square terms in the equation this is a quadratic equation, so we can use the method above.
Do:
$2x^2+4x+2$2x2+4x+2  $=$=  $x^24x13$x2−4x−13 

$x^2+8x+15=0$x2+8x+15=0  $=$=  $0$0 
Gathering all of the terms on the left hand side and simplifying 
$\left(x+5\right)\left(x+3\right)$(x+5)(x+3)  $=$=  $0$0 
Factorising the quadratic expression on the left hand side Note that $5+3=8$5+3=8 and $5\times3=15$5×3=15 
Now we use the null factor law to split this quadratic equation into two linear equations.
$x+5$x+5  $=$=  $0$0 
Using one factor 
$x$x  $=$=  $5$−5 
Solving for $x$x 
$x+3$x+3  $=$=  $0$0 
Using the other factor 
$x$x  $=$=  $3$−3 
Solving for $x$x 
So the solutions are $x=5$x=−5 and $x=3$x=−3.
We can solve many quadratic equations by using this method:
Solve $2y6y^2=0$2y−6y2=0 for $y$y.
Enter each solution on the same line, separated by a comma.
Solve $x^2=13x+114$x2=13x+114 for $x$x.
Enter each solution on the same line, separated by a comma.
Solve $x^2+13x+42=0$x2+13x+42=0 for $x$x.
Enter each solution on the same line, separated by a comma.
Solve quadratic equations for real roots.