# Problem of the Week

## Updated at Jun 17, 2024 3:09 PM

To get more practice in equation, we brought you this problem of the week:

How can we solve the equation $$2+{x}^{2}-\frac{4}{5}x=\frac{11}{5}$$?

Check out the solution below!

$2+{x}^{2}-\frac{4}{5}x=\frac{11}{5}$

 1 Simplify  $$\frac{4}{5}x$$  to  $$\frac{4x}{5}$$.$2+{x}^{2}-\frac{4x}{5}=\frac{11}{5}$2 Multiply both sides by $$5$$.$10+5{x}^{2}-4x=11$3 Move all terms to one side.$10+5{x}^{2}-4x-11=0$4 Simplify  $$10+5{x}^{2}-4x-11$$  to  $$-1+5{x}^{2}-4x$$.$-1+5{x}^{2}-4x=0$5 Split the second term in $$-1+5{x}^{2}-4x$$ into two terms.1 Multiply the coefficient of the first term by the constant term.$5\times -1=-5$2 Ask: Which two numbers add up to $$-4$$ and multiply to $$-5$$?$$1$$ and $$-5$$3 Split $$-4x$$ as the sum of $$x$$ and $$-5x$$.$5{x}^{2}+x-5x-1$To get access to all 'How?' and 'Why?' steps, join Cymath Plus!$5{x}^{2}+x-5x-1=0$6 Factor out common terms in the first two terms, then in the last two terms.$x(5x+1)-(5x+1)=0$7 Factor out the common term $$5x+1$$.$(5x+1)(x-1)=0$8 Solve for $$x$$.1 Ask: When will $$(5x+1)(x-1)$$ equal zero?When $$5x+1=0$$ or $$x-1=0$$2 Solve each of the 2 equations above.$x=-\frac{1}{5},1$To get access to all 'How?' and 'Why?' steps, join Cymath Plus!$x=-\frac{1}{5},1$DoneDecimal Form: -0.2, 1x=-1/5,1