If it's not what You are looking for type in the equation solver your own equation and let us solve it.
(2x)^2-100=0
a = 2; b = 0; c = -100;
Δ = b2-4ac
Δ = 02-4·2·(-100)
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*2}=\frac{0-20\sqrt{2}}{4} =-\frac{20\sqrt{2}}{4} =-5\sqrt{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20\sqrt{2}}{2*2}=\frac{0+20\sqrt{2}}{4} =\frac{20\sqrt{2}}{4} =5\sqrt{2} $
| j-63/8=4 | | 5z+5/4=6z+7/4 | | 4(k+3)-2(k-3)=38 | | 2/3(6x+1)=6+1/3(12x+8) | | 6-x÷5=8 | | f/0.2.5=16 | | 2/3x-11=-9 | | 0.5n+7+n=20 | | 2(0.5x+1)=-3(2x-1)+4(2x+1) | | 30x+78=30x+37 | | s/7+19=24 | | 7x+23x-6=6(5x+5) | | -6=0+b | | 4/x+1=5/3x+1 | | x=0.5(x-4+2x-3) | | 8x-128/5=-24x+4+3/5 | | 14+m/10=2 | | 5(4+j)=100 | | 8x-3=5(×+1)+3x-8 | | (2x+3)(x-4)-(x^2x+6)=0 | | 0=r^2-4r-32 | | 1.6x-2.88=3.4x | | -3b+19=1+9b+18 | | 8+x10=5 | | -2(x-5)=40+x | | 7x+13x-7=4(5x+8) | | 39=4y+3 | | 160=4(9)(4)+(2)(4)(x) | | 46x+2,704x-80=55(50x+61) | | 18=8y+2 | | 8p-15-5p=21p-3 | | 3c–8=1 |