Solve 315=x+x^2/20

$$315=x+ \frac{ { x }^{ 2 } }{ 20 }$$

Solve for x

$$x=-90$$

$$x=70$$

Steps Using Factoring

Multiply both sides of the equation by 20.

$$6300=20x+x^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$20x+x^{2}=6300$$

Subtract 6300 from both sides.

$$20x+x^{2}-6300=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$x^{2}+20x-6300=0$$

To solve the equation, factor x^{2}+20x-6300 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=20 ab=-6300$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -6300.

$$-1,6300 -2,3150 -3,2100 -4,1575 -5,1260 -6,1050 -7,900 -9,700 -10,630 -12,525 -14,450 -15,420 -18,350 -20,315 -21,300 -25,252 -28,225 -30,210 -35,180 -36,175 -42,150 -45,140 -50,126 -60,105 -63,100 -70,90 -75,84$$

Calculate the sum for each pair.

$$-1+6300=6299 -2+3150=3148 -3+2100=2097 -4+1575=1571 -5+1260=1255 -6+1050=1044 -7+900=893 -9+700=691 -10+630=620 -12+525=513 -14+450=436 -15+420=405 -18+350=332 -20+315=295 -21+300=279 -25+252=227 -28+225=197 -30+210=180 -35+180=145 -36+175=139 -42+150=108 -45+140=95 -50+126=76 -60+105=45 -63+100=37 -70+90=20 -75+84=9$$

The solution is the pair that gives sum 20.

$$a=-70 b=90$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-70\right)\left(x+90\right)$$

To find equation solutions, solve x-70=0 and x+90=0.

$$x=70 x=-90$$

Steps Using Factoring By Grouping

Multiply both sides of the equation by 20.

$$6300=20x+x^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$20x+x^{2}=6300$$

Subtract 6300 from both sides.

$$20x+x^{2}-6300=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$x^{2}+20x-6300=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-6300. To find a and b, set up a system to be solved.

$$a+b=20 ab=1\left(-6300\right)=-6300$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -6300.

$$-1,6300 -2,3150 -3,2100 -4,1575 -5,1260 -6,1050 -7,900 -9,700 -10,630 -12,525 -14,450 -15,420 -18,350 -20,315 -21,300 -25,252 -28,225 -30,210 -35,180 -36,175 -42,150 -45,140 -50,126 -60,105 -63,100 -70,90 -75,84$$

Calculate the sum for each pair.

$$-1+6300=6299 -2+3150=3148 -3+2100=2097 -4+1575=1571 -5+1260=1255 -6+1050=1044 -7+900=893 -9+700=691 -10+630=620 -12+525=513 -14+450=436 -15+420=405 -18+350=332 -20+315=295 -21+300=279 -25+252=227 -28+225=197 -30+210=180 -35+180=145 -36+175=139 -42+150=108 -45+140=95 -50+126=76 -60+105=45 -63+100=37 -70+90=20 -75+84=9$$

The solution is the pair that gives sum 20.

$$a=-70 b=90$$

Rewrite x^{2}+20x-6300 as \left(x^{2}-70x\right)+\left(90x-6300\right).

$$\left(x^{2}-70x\right)+\left(90x-6300\right)$$

Factor out x in the first and 90 in the second group.

$$x\left(x-70\right)+90\left(x-70\right)$$

Factor out common term x-70 by using distributive property.

$$\left(x-70\right)\left(x+90\right)$$

To find equation solutions, solve x-70=0 and x+90=0.

$$x=70 x=-90$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 20.

$$6300=20x+x^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$20x+x^{2}=6300$$

Subtract 6300 from both sides.

$$20x+x^{2}-6300=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$x^{2}+20x-6300=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 20 for b, and -6300 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-20±\sqrt{20^{2}-4\left(-6300\right)}}{2}$$

Square 20.

$$x=\frac{-20±\sqrt{400-4\left(-6300\right)}}{2}$$

Multiply -4 times -6300.

$$x=\frac{-20±\sqrt{400+25200}}{2}$$

Add 400 to 25200.

$$x=\frac{-20±\sqrt{25600}}{2}$$

Take the square root of 25600.

$$x=\frac{-20±160}{2}$$

Now solve the equation x=\frac{-20±160}{2} when ± is plus. Add -20 to 160.

$$x=\frac{140}{2}$$

Divide 140 by 2.

$$x=70$$

Now solve the equation x=\frac{-20±160}{2} when ± is minus. Subtract 160 from -20.

$$x=-\frac{180}{2}$$

Divide -180 by 2.

$$x=-90$$

The equation is now solved.

$$x=70 x=-90$$

Steps for Completing the Square

Multiply both sides of the equation by 20.

$$6300=20x+x^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$20x+x^{2}=6300$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$x^{2}+20x=6300$$

Divide 20, the coefficient of the x term, by 2 to get 10. Then add the square of 10 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+20x+10^{2}=6300+10^{2}$$

Square 10.

$$x^{2}+20x+100=6300+100$$

Add 6300 to 100.

$$x^{2}+20x+100=6400$$

Factor x^{2}+20x+100. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+10\right)^{2}=6400$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+10\right)^{2}}=\sqrt{6400}$$

Simplify.

$$x+10=80 x+10=-80$$

Subtract 10 from both sides of the equation.

$$x=70 x=-90$$