If it's not what You are looking for type in the equation solver your own equation and let us solve it.
20x^2=10
We move all terms to the left:
20x^2-(10)=0
a = 20; b = 0; c = -10;
Δ = b2-4ac
Δ = 02-4·20·(-10)
Δ = 800
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{800}=\sqrt{400*2}=\sqrt{400}*\sqrt{2}=20\sqrt{2}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20\sqrt{2}}{2*20}=\frac{0-20\sqrt{2}}{40} =-\frac{20\sqrt{2}}{40} =-\frac{\sqrt{2}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20\sqrt{2}}{2*20}=\frac{0+20\sqrt{2}}{40} =\frac{20\sqrt{2}}{40} =\frac{\sqrt{2}}{2} $
| 2=1^x | | 1/2x+200=5/6x | | (v-3)/4=2 | | 1.7=1.05^m | | 60=-2x+14 | | 3x+5=6x5 | | 1.5=1.05^n | | 6(3x–5)–5(2x+7)=25 | | 14-4x=22 | | 3x^2+6x-35=0 | | 5x-8+x=3x+5x+12 | | 13x-13=39 | | 7x-3x=-6(x+20)+5(-3-3x) | | 20x-128=x | | t^2-6t-144=0 | | 2x-2x-x=-4 | | 9x+1225=81+35x | | -15x=-5(3x+) | | f(-3)=2(-3)^2+8(-3)-4 | | 10+0.8r=20+0.4r | | 4-4x+3x=-2 | | y=1.50+75 | | 8x+126=3x+9 | | -17+3v=9(4v+6) | | 27=4x+x-3 | | y/2-1.5=1.7 | | 2x( | | 6m+4/24=5m+8/27 | | 5/3=n/11 | | 2n=n-1 | | 6a+5+3a+8=180 | | 4x-24+2=6x+6 |